# point of tangency of a circle formula

The tangent to the circle at the point $$(2;2)$$ is perpendicular to the radius, so $$m \times m_{\text{tangent}} = -1$$. Sketch the circle and the straight line on the same system of axes. That would be the tiny trail the circlemakers walked along to get to the spot in the field where they started forming their crop circle. The product of the gradient of the radius and the gradient of the tangent line is equal to $$-\text{1}$$. The tangent is perpendicular to the radius, therefore $$m \times m_{\bot} = -1$$. In geometry, a tangent of a circle is a straight line that touches the circle at exactly one point, never entering the circle’s interior. The tangents to the circle, parallel to the line $$y = \frac{1}{2}x + 1$$, must have a gradient of $$\frac{1}{2}$$. In other words, we can say that the lines that intersect the circles exactly in one single point are Tangents. to personalise content to better meet the needs of our users. &= \sqrt{(6)^{2} + (-12)^2} \\ A circle with centre $$C(a;b)$$ and a radius of $$r$$ units is shown in the diagram above. In simple words, we can say that the lines that intersect the circle exactly in one single point are tangents. The equation of the tangent at point $$A$$ is $$y = \frac{1}{2}x + 11$$ and the equation of the tangent at point $$B$$ is $$y = \frac{1}{2}x - 9$$. &= \sqrt{(2 -(-10))^{2} + (4 - 10)^2} \\ The equation of the tangent to the circle is $$y = 7 x + 19$$. where r is the circle’s radius. Solution: Slopes and intersections of common tangents to the circles must satisfy tangency condition of both circles.Therefore, values for slopes m and intersections c we calculate from the system of equations, \begin{align*} The equation of the tangent to the circle at $$F$$ is $$y = - \frac{1}{4}x + \frac{9}{2}$$. So, if you have a graph with curves, like a parabola, it can have points of tangency as well. So the circle's center is at the origin with a radius of about 4.9. The equation of tangent to the circle {x^2} + {y^2} A chord and a secant connect only two points on the circle. More precisely, a straight line is said to be a tangent of a curve y = f(x) at a point x = c if the line passes through the point (c, f(c)) on the curve and has slope f '(c), where f ' is the derivative of f. A circle can have a: Here is a crop circle that shows the flattened crop, a center point, a radius, a secant, a chord, and a diameter: [insert cartoon crop circle as described and add a tangent line segment FO at the 2-o'clock position; label the circle's center U]. The points will be where the circle's equation = the tangent's … The coordinates of the centre of the circle are $$(-4;-8)$$. Where it touches the line, the equation of the circle equals the equation of the line. It is a line through a pair of infinitely close points on the circle. A circle has a center, which is that point in the middle and provides the name of the circle. Tangents, of course, also allude to writing or speaking that diverges from the topic, as when a writer goes off on a tangent and points out that most farmers do not like having their crops stomped down by vandals from this or any other world. From the equation, determine the coordinates of the centre of the circle $$(a;b)$$. This perpendicular line will cut the circle at $$A$$ and $$B$$. The tangent to the circle at the point $$(5;-5)$$ is perpendicular to the radius of the circle to that same point: $$m \times m_{\bot} = -1$$. To determine the coordinates of $$A$$ and $$B$$, we substitute the straight line $$y = - 2x + 1$$ into the equation of the circle and solve for $$x$$: This gives the points $$A(-4;9)$$ and $$B(4;-7)$$. by this license. &= \left( \frac{-2}{2}; \frac{2}{2} \right) \\ This forms a crop circle nest of seven circles, with each outer circle touching exactly three other circles, and the original center circle touching exactly six circles: Three theorems (that do not, alas, explain crop circles) are connected to tangents. \end{align*}. m_r & = \frac{y_1 - y_0}{x_1 - x_0} \\ This means a circle is not all the space inside it; it is the curved line around a point that closes in a space. Setting each equal to 0 then setting them equal to each other might help. Write down the equation of a straight line and substitute $$m = 7$$ and $$(-2;5)$$. Determine the coordinates of $$S$$, the point where the two tangents intersect. Finally we convert that angle to degrees with the 180 / π part. Lines and line segments are not the only geometric figures that can form tangents. Here, the list of the tangent to the circle equation is given below: 1. Recall that the equation of the tangent to this circle will be y = mx ± a$$\small \sqrt{1+m^2}$$ . & \\ &= \sqrt{180} PS &= \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^2} \\ Solution : Equation of the line 3x + 4y − p = 0. D(x; y) is a point on the circumference and the equation of the circle is: (x − a)2 + (y − b)2 = r2 A tangent is a straight line that touches the circumference of a circle at … The point P is called the point … Method 1. Solve these 4 equations simultaneously to find the 4 unknowns (c,d), and (e,f). Find the radius r of O. Tangent at point P is the limiting position of a secant PQ when Q tends to P along the circle. Determine the coordinates of $$H$$, the mid-point of chord $$PQ$$. Draw $$PT$$ and extend the line so that is cuts the positive $$x$$-axis. Suppose it is 7 units. If (2,10) is a point on the tangent, how do I find the point of tangency on the circle? Here we list the equations of tangent and normal for different forms of a circle and also list the condition of tangency for the line to a circle. This formula works because dy / dx gives the slope of the line created by the movement of the circle across the plane. To find the equation of the second parallel tangent: All Siyavula textbook content made available on this site is released under the terms of a You can also surround your first crop circle with six circles of the same diameter as the first. The tangent of a circle is perpendicular to the radius, therefore we can write: Substitute $$m_{Q} = - \frac{1}{2}$$ and $$Q(2;4)$$ into the equation of a straight line. \end{align*}. The tangent to a circle is perpendicular to the radius at the point of tangency. Plugging the points into y = x3 gives you the three points: (–1.539, –3.645), (–0.335, –0.038), and (0.250, 0.016). The required equation will be x(4) + y(-3) = 25, or 4x – 3y = 25. In the circle O , P T ↔ is a tangent and O P ¯ is the radius. Notice that the line passes through the centre of the circle. &= \sqrt{(-6)^{2} + (-6)^2} \\ c 2 = a 2 (1 + m 2) p 2 /16 = 16 (1 + 9/16) p 2 /16 = 16 (25/16) p 2 /16 = 25. p 2 = 25(16) p = ± 20. QS &= \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^2} \\ \begin{align*} Make a conjecture about the angle between the radius and the tangent to a circle at a point on the circle. The equations of the tangents to the circle are $$y = - \frac{3}{4}x - \frac{25}{4}$$ and $$y = \frac{4}{3}x + \frac{25}{3}$$. We do not know the slope. $$C(-4;8)$$ is the centre of the circle passing through $$H(2;-2)$$ and $$Q(-10;m)$$. Solved: In the diagram, point P is a point of tangency. radius (the distance from the center to the circle), chord (a line segment from the circle to another point on the circle without going through the center), secant (a line passing through two points of the circle), diameter (a chord passing through the center). Plot the point $$P(0;5)$$. United States. At this point, you can use the formula, \\ m \angle MJK= \frac{1}{2} \cdot 144 ^{\circ} \\ m \angle ... Back to Circle Formulas Next to Arcs and Angles. the centre of the circle $$(a;b) = (8;-7)$$, a point on the circumference of the circle $$(x_1;y_1) = (5;-5)$$, the equation for the circle $$\left(x + 4\right)^{2} + \left(y + 8\right)^{2} = 136$$, a point on the circumference of the circle $$(x_1;y_1) = (2;2)$$, the centre of the circle $$C(a;b) = (1;5)$$, a point on the circumference of the circle $$H(-2;1)$$, the equation for the tangent to the circle in the form $$y = mx + c$$, the equation for the circle $$\left(x - 4\right)^{2} + \left(y + 5\right)^{2} = 5$$, a point on the circumference of the circle $$P(2;-4)$$, the equation of the tangent in the form $$y = mx + c$$. Tangent to a circle: Let P be a point on circle and let PQ be secant. Several theorems are related to this because it plays a significant role in geometrical constructionsand proofs. This line runs parallel to the line y=5x+7. Substitute the straight line $$y = x + 4$$ into the equation of the circle and solve for $$x$$: This gives the points $$P(-5;-1)$$ and $$Q(1;5)$$. The second theorem is called the Two Tangent Theorem. This gives the point $$S \left( - 10;10 \right)$$. \begin{align*} Crop circles almost always "appear" very close to roads and show some signs of tangents, which is why most researchers say they are made by human pranksters. I need to find the points of tangency between the line y=5x+b and the circle. The line joining the centre of the circle to this point is parallel to the vector. m_{PQ} \times m_{OM} &= - 1 \\ The intersection point of the outer tangents lines is: (-3.67 ,4.33) Note: r 0 should be the bigger radius in the equation of the intersection. &= \sqrt{180} From the given equation of $$PQ$$, we know that $$m_{PQ} = 1$$. We can also talk about points of tangency on curves. Use the distance formula to determine the length of the radius: Write down the general equation of a circle and substitute $$r$$ and $$H(2;-2)$$: The equation of the circle is $$\left(x + 4\right)^{2} + \left(y - 8\right)^{2} = 136$$. The gradient for the tangent is $$m_{\text{tangent}} = - \frac{3}{5}$$. The coordinates of the centre of the circle are $$(a;b) = (4;-5)$$. &= \sqrt{(-4 -2)^{2} + (-2-4 )^2} \\ \end{align*} Circle centered at any point (h, k), ( x – h) 2 + ( y – k) 2 = r2. &= \left( \frac{-4 + 2}{2}; \frac{-2 + 4}{2} \right) \\ Determine the equation of the circle and write it in the form $(x - a)^{2} + (y - b)^{2} = r^{2}$. The centre of the circle is $$(-3;1)$$ and the radius is $$\sqrt{17}$$ units. circumference (the distance around the circle itself. Determine the coordinates of $$M$$, the mid-point of chord $$PQ$$. x 2 + y 2 = r 2. The angle T is a right angle because the radius is perpendicular to the tangent at the point of tangency, AT¯ ⊥ TP↔. Points of tangency do not happen just on circles. At the point of tangency, a tangent is perpendicular to the radius. Two-Tangent Theorem: When two segments are drawn tangent to a circle from the same point outside the circle, the segments are equal in length. Let the gradient of the tangent at $$P$$ be $$m_{P}$$. &= 6\sqrt{2} Find a tutor locally or online. Similarly, $$H$$ must have a positive $$y$$-coordinate, therefore we take the positive of the square root. [insert diagram of circle A with tangent LI perpendicular to radius AL and secant EN that, beyond the circle, also intersects Point I]. We think you are located in 1-to-1 tailored lessons, flexible scheduling. Let the gradient of the tangent at $$Q$$ be $$m_{Q}$$. The solution shows that $$y = -2$$ or $$y = 18$$. Determine the gradient of the radius $$OT$$. Equation (4) represents the fact that the distance between the point of tangency and the center of circle 2 is r2, or (f-b)^2 + (e-a)^2 = r2^2. Here we have circle A where AT¯ is the radius and TP↔ is the tangent to the circle. Determine the gradient of the radius: $m_{CD} = \frac{y_{2} - y_{1}}{x_{2}- x_{1}}$, The radius is perpendicular to the tangent of the circle at a point $$D$$ so: $m_{AB} = - \frac{1}{m_{CD}}$, Write down the gradient-point form of a straight line equation and substitute $$m_{AB}$$ and the coordinates of $$D$$. The tangent to a circle equation x2+ y2+2gx+2fy+c =0 at (x1, y1) is xx1+yy1+g(x+x1)+f(y +y1)+c =0 1.3. The gradient of the radius is $$m = - \frac{2}{3}$$. On a suitable system of axes, draw the circle $$x^{2} + y^{2} = 20$$ with centre at $$O(0;0)$$. The equation for the tangent to the circle at the point $$Q$$ is: The straight line $$y = x + 2$$ cuts the circle $$x^{2} + y^{2} = 20$$ at $$P$$ and $$Q$$. We’ll use the point form once again. After working your way through this lesson and video, you will learn to: Get better grades with tutoring from top-rated private tutors. The equations of the tangents are $$y = -5x - 26$$ and $$y = - \frac{1}{5}x + \frac{26}{5}$$. Get help fast. The tangent lines to circles form the subject of several theorems and play an important role in many geometrical constructions and proofs. Join thousands of learners improving their maths marks online with Siyavula Practice. \end{align*}. That distance is known as the radius of the circle. The point where a tangent touches the circle is known as the point of tangency. &= \frac{6}{6} \\ The same reciprocal relation exists between a point P outside the circle and the secant line joining its two points of tangency. Equate the two linear equations and solve for $$x$$: This gives the point $$S \left( - \frac{13}{2}; \frac{13}{2} \right)$$. Therefore the equations of the tangents to the circle are $$y = -2x - 10$$ and $$y = - \frac{1}{2}x + 5$$. We wil… \end{align*}. If $$O$$ is the centre of the circle, show that $$PQ \perp OH$$. How do we find the length of AP¯? Consider $$\triangle GFO$$ and apply the theorem of Pythagoras: Note: from the sketch we see that $$F$$ must have a negative $$y$$-coordinate, therefore we take the negative of the square root. Is this correct? Substitute the straight line $$y = x + 2$$ into the equation of the circle and solve for $$x$$: This gives the points $$P(-4;-2)$$ and $$Q(2;4)$$. The radius of the circle $$CD$$ is perpendicular to the tangent $$AB$$ at the point of contact $$D$$. Example: At intersections of a line x-5y + 6 = 0 and the circle x 2 + y 2-4x + 2y -8 = 0 drown are tangents, find the area of the triangle formed by the line and the tangents. Determine the gradient of the tangent to the circle at the point $$(2;2)$$. We need to show that the product of the two gradients is equal to $$-\text{1}$$. The radius is perpendicular to the tangent, so $$m \times m_{\bot} = -1$$. Point Of Tangency To A Curve. We are interested in ﬁnding the equations of these tangent lines (i.e., the lines which pass through exactly one point of the circle, and pass through (5;3)). Substitute the $$Q(-10;m)$$ and solve for the $$m$$ value. Complete the sentence: the product of the $$\ldots \ldots$$ of the radius and the gradient of the $$\ldots \ldots$$ is equal to $$\ldots \ldots$$. &= \sqrt{36 + 36} \\ Let the two tangents from $$G$$ touch the circle at $$F$$ and $$H$$. Calculate the coordinates of $$P$$ and $$Q$$. \begin{align*} m_{OM} &= \frac{1 - 0}{-1 - 0} \\ \therefore PQ & \perp OM A tangent to a circle is a straight line that touches the circle at one point, called the point of tangency. Determine the equations of the tangents to the circle at $$P$$ and $$Q$$. Specifically, my problem deals with a circle of the equation x^2+y^2=24 and the point on the tangent being (2,10). where ( … Example: Find the outer intersection point of the circles: (r 0) (x − 3) 2 + (y + 5) 2 = 4 2 (r 1) (x + 2) 2 + (y − 2) 2 = 1 2. Make $$y$$ the subject of the equation. In our crop circle U, if we look carefully, we can see a tangent line off to the right, line segment FO. A tangent connects with only one point on a circle. A tangent to a circle is a straight line that touches the circle at one point, called the point of tangency. Example 3 : Find the value of p so that the line 3x + 4y − p = 0 is a tangent to x 2 + y 2 − 64 = 0. The tangent of a circle is perpendicular to the radius, therefore we can write: Substitute $$m_{P} = - 2$$ and $$P(-4;-2)$$ into the equation of a straight line. Here is a crop circle with three little crop circles tangential to it: [insert cartoon drawing of a crop circle ringed by three smaller, tangential crop circles]. This gives the points $$F(-3;-4)$$ and $$H(-4;3)$$. The gradient for this radius is $$m = \frac{5}{3}$$. The equation of the tangent to the circle is. Tangent to a Circle A tangent to a circle is a straight line which touches the circle at only one point. The tangent to a circle equation x2+ y2=a2 at (x1, y1) isxx1+yy1= a2 1.2. Leibniz defined it as the line through a pair of infinitely close points on the curve. We already snuck one past you, like so many crop circlemakers skulking along a tangent path: a tangent is perpendicular to a radius. Point of tangency is the point where the tangent touches the circle. To do that, the tangent must also be at a right angle to a radius (or diameter) that intersects that same point. This also works if we use the slope of the surface. The two vectors are orthogonal, so their dot product is zero: w = ( 1 2) (it has gradient 2 ). Given a circle with the central coordinates $$(a;b) = (-9;6)$$. From the sketch we see that there are two possible tangents. If $$O$$ is the centre of the circle, show that $$PQ \perp OM$$. The equation for the tangent to the circle at the point $$H$$ is: Given the point $$P(2;-4)$$ on the circle $$\left(x - 4\right)^{2} + \left(y + 5\right)^{2} = 5$$. Equation of the circle x 2 + y 2 = 64. The tangent at $$P$$, $$y = -2x - 10$$, is parallel to $$y = - 2x + 4$$. Determine the gradient of the radius $$OQ$$: Substitute $$m_{Q} = - \frac{1}{5}$$ and $$Q(1;5)$$ into the equation of a straight line. The two circles could be nested (one inside the other) or adjacent. &= \sqrt{36 \cdot 2} \\ &= \sqrt{(-4 -(-10))^{2} + (-2 - 10)^2} \\ Because equations (3) and (4) are quadratic, there will be as many as 4 solutions, as shown in the picture. Circles are the set of all points a given distance from a point. This point is called the point of tangency. In geometry, a circle is a closed curve formed by a set of points on a plane that are the same distance from its center O. &= - 1 \\ We derive the equation of tangent line for a circle with radius r. For simplicity, we chose for the origin the centre of the circle, when the points (x, y) of the circle satisfy the equation. Measure the angle between $$OS$$ and the tangent line at $$S$$. Identify and recognize a tangent of a circle, Demonstrate how circles can be tangent to other circles, Recall and explain three theorems related to tangents. Determine the equation of the tangent to the circle at the point $$(-2;5)$$. In the following diagram: If AB and AC are two tangents to a circle centered at O, then: the tangents to the circle from the external point A are equal, &= 1 \\ How to determine the equation of a tangent: Determine the equation of the tangent to the circle $$x^{2} + y^{2} - 2y + 6x - 7 = 0$$ at the point $$F(-2;5)$$. I need to find the points of tangency on a circle (x^2+y^2=100) and a line y=5x+b the only thing I know about b is that it is negative. Example 2 Find the equation of the tangent to the circle x 2 + y 2 – 2x – 6y – 15 = 0 at the point (5, 6). We have already shown that $$PQ$$ is perpendicular to $$OH$$, so we expect the gradient of the line through $$S$$, $$H$$ and $$O$$ to be $$-\text{1}$$. $$D(x;y)$$ is a point on the circumference and the equation of the circle is: A tangent is a straight line that touches the circumference of a circle at only one place. The key is to ﬁnd the points of tangency, labeled A 1 and A 2 in the next ﬁgure. This means we can use the Pythagorean Theorem to solve for AP¯. We need to show that there is a constant gradient between any two of the three points. Here a 2 = 16, m = −3/4, c = p/4. Notice that the diameter connects with the center point and two points on the circle. Determine the equations of the two tangents to the circle, both parallel to the line $$y + 2x = 4$$. then the equation of the circle is (x-12)^2+ (y-10)^2=49, the radius squared. Solution This one is similar to the previous problem, but applied to the general equation of the circle. & \\ &= \sqrt{36 + 144} \\ \begin{align*} Determine the gradient of the radius $$OP$$: The tangent of a circle is perpendicular to the radius, therefore we can write: Substitute $$m_{P} = - 5$$ and $$P(-5;-1)$$ into the equation of a straight line. Apart from the stuff given in this section "Find the equation of the tangent to the circle at the point", if you need any other stuff in math, please use our google custom search here. Learn faster with a math tutor. Once we have the slope, we take the inverse tangent (arctan) of it which gives its angle in radians. So, you find that the point of tangency is (2, 8); the equation of tangent line is y = 12 x – 16; and the points of normalcy are approximately (–1.539, –3.645), (–0.335, –0.038), and (0.250, 0.016). m_{PQ} &= \frac{4 - (-2)}{2 - (-4)} \\ Below, we have the graph of y = x^2. To determine the coordinates of $$A$$ and $$B$$, we must find the equation of the line perpendicular to $$y = \frac{1}{2}x + 1$$ and passing through the centre of the circle. & = - \frac{1}{7} equation of tangent of circle. After having gone through the stuff given above, we hope that the students would have understood "Find the equation of the tangent to the circle at the point ". Popular pages @ mathwarehouse.com . Condition of Tangency: The line y = mx + c touches the circle x² + y² = a² if the length of the intercepts is zero i.e., c = ± a √(1 + m²). v = ( a − 3 b − 4) The line y = 2 x + 3 is parallel to the vector. Get better grades with tutoring from top-rated professional tutors. Though we may not have solved the mystery of crop circles, you now are able to identify the parts of a circle, identify and recognize a tangent of a circle, demonstrate how circles can be tangent to other circles, and recall and explain three theorems related to tangents of circles. The straight line $$y = x + 4$$ cuts the circle $$x^{2} + y^{2} = 26$$ at $$P$$ and $$Q$$. Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there! The Tangent Secant Theorem explains a relationship between a tangent and a secant of the same circle. The gradient for the tangent is $$m_{\bot} = \frac{3}{2}$$. This means that AT¯ is perpendicular to TP↔. Determine the equation of the tangent to the circle with centre $$C$$ at point $$H$$. EF is a tangent to the circle and the point of tangency is H. Tangents From The Same External Point. Label points $$P$$ and $$Q$$. One circle can be tangent to another, simply by sharing a single point. Therefore $$S$$, $$H$$ and $$O$$ all lie on the line $$y=-x$$. We won’t establish any formula here, but I’ll illustrate two different methods, first using the slope form and the other using the condition of tangency. Local and online. 1.1. Write down the gradient-point form of a straight line equation and substitute $$m = - \frac{1}{4}$$ and $$F(-2;5)$$. Determine the gradient of the tangent to the circle at the point $$(5;-5)$$. (1) Let the point of tangency be (x 0, y 0). Want to see the math tutors near you? With Point I common to both tangent LI and secant EN, we can establish the following equation: Though it may sound like the sorcery of aliens, that formula means the square of the length of the tangent segment is equal to the product of the secant length beyond the circle times the length of the whole secant. $y - y_{1} = m(x - x_{1})$. Determine the equation of the tangent to the circle at point $$Q$$. \end{align*}. Determine the equations of the tangents to the circle $$x^{2} + y^{2} = 25$$, from the point $$G(-7;-1)$$ outside the circle. \begin{align*} Plot the point $$S(2;-4)$$ and join $$OS$$. Show that $$S$$, $$H$$ and $$O$$ are on a straight line. A circle with centre $$(8;-7)$$ and the point $$(5;-5)$$ on the circle are given. We use this information to present the correct curriculum and Example: Find equations of the common tangents to circles x 2 + y 2 = 13 and (x + 2) 2 + (y + 10) 2 = 117. At the point of tangency, the tangent of the circle is perpendicular to the radius. A tangent is a line (or line segment) that intersects a circle at exactly one point. The line that joins two infinitely close points from a point on the circle is a Tangent. The tangent to a circle equation x2+ y2=a2 for a line y = mx +c is y = mx ± a √[1+ m2] From the graph we see that the $$y$$-coordinate of $$Q$$ must be positive, therefore $$Q(-10;18)$$. Tangent to a Circle. Only one tangent can be at a point to circle. Let the gradient of the tangent line be $$m$$. &= \left( -1; 1 \right) The tangent line $$AB$$ touches the circle at $$D$$. &= \sqrt{144 + 36} \\ Here are the circle equations: Circle centered at the origin, (0, 0), x2 + y2 = r2. If a point P is exterior to a circle with center O, and if the tangent lines from P touch the circle at points T and S, then ∠TPS and ∠TOS are supplementary (sum to 180°). Points of tangency do not happen just on circles. The condition for the tangency is c 2 = a 2 (1 + m 2) . We can also talk about points of tangency on curves. The word "tangent" comes from a Latin term meaning "to touch," because a tangent just barely touches a circle. The points on the circle can be calculated when you know the equation for the tangent lines. Equation of the two circles given by: (x − a) 2 + (y − b) 2 = r 0 2 (x − c) 2 + (y − d) 2 = r 1 2. Let's look at an example of that situation. Determine the gradient of the radius. Embedded videos, simulations and presentations from external sources are not necessarily covered Determine the equations of the tangents to the circle $$x^{2} + (y - 1)^{2} = 80$$, given that both are parallel to the line $$y = \frac{1}{2}x + 1$$. Let's try an example where AT¯ = 5 and TP↔ = 12. PQ &= \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^2} \\ &= \sqrt{(12)^{2} + (-6)^2} \\ Given the equation of the circle: $$\left(x + 4\right)^{2} + \left(y + 8\right)^{2} = 136$$. At the point of tangency, the tangent of the circle is perpendicular to the radius. Creative Commons Attribution License. Find the gradient of the radius at the point $$(2;2)$$ on the circle. Here we have circle A A where ¯¯¯¯¯ ¯AT A T ¯ is the radius and ←→ T P T P ↔ is the tangent to the circle. 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