Table of Contents

## Tangents to a Circle:

You know that if a line intersects a circle at two coincident points, then the line is called a tangent to the circle at the point of contact.

In the figure, the line PQ intersects the circle at P and Q. If the line PQ is now rotated about P in the clockwise direction as shown in the figure, then ultimately a position is reached when Q coincides with P. At this stage, the line becomes tangent to the circle at P. The point P is known as the point of tangency or the point of contact.

### Equation of the Tangent to the Circle x^{2} + y^{2} = r^{2} at (x_{1}, y_{1}):

Let P be the point (x_{1}, y_{1}) on the circle x^{2} + y^{2} = r^{2}. We take a neighboring point Q (x_{2}, y_{2}) on the circle as shown in figure.Thus, x _{1}^{2} + y_{1}^{2} = r^{2} ……….(i)and x _{2}^{2} + y_{2}^{2} = r^{2} ……….(ii)Equation to PQ is, y – y _{1} = [ (y_{2} – y_{1}) / (x_{2} – x_{1}) ] (x – x_{1}) ……….(iii)From (i) and (ii), we get x _{2}^{2} – x_{1}^{2} + y_{2}^{2} – y_{1}^{2} = 0⇒ (x _{2} + x_{1}) (x_{2} – x_{1}) = – (y_{2} + y_{1}) (y_{2} – y_{1})⇒ (y _{2} – y_{1}) / (x_{2} – x_{1}) = – (x_{1} + x_{2}) / (y_{1} + y_{2})∴ Equation to PQ becomes, y – y _{1} = – [ (x_{1} + x_{2}) / (y_{1} + y_{2}) ] (x – x_{1})As Q → P, x_{2} → x_{1,} and y_{2} → y_{2}, and the line PQ becomes a tangent to the circle at P. Thus, the equation to the tangent to the circle x^{2} + y^{2} = r^{2} at P isy – y _{1} = – (2x_{1}/2y_{1}) (x – x_{1})⇒ x _{1} (x – x_{1}) + y_{1} (y – y_{1}) = 0⇒ xx _{1} + yy_{1} = x^{2}_{1} + y^{2}_{1} = r^{2} (∵ x^{2} + y^{2} = r^{2}) |

### Equation of the Tangent to the Circle x^{2} + y^{2} + 2gx + 2fy + c= 0 at the point (x_{1}, y_{1}):

Let P be the point (x_{1}, y_{1}) on the circle x^{2} + y^{2} + 2gx + 2fy + c= 0.We take a neighboring point Q (x _{2}, y_{2}) on the circle as shown in the figure.As P and Q lie on the circle, we have x _{1}^{2} + y_{1}^{2} + 2gx_{1} + 2fy_{1} + c= 0 …………(i)and x _{2}^{2} + y_{2}^{2} + 2gx_{2} + 2fy_{2} + c= 0 …………(ii)∴ Subtracting (i) from (ii), we get (x _{2}^{2} – x_{1}^{2}) + (y_{2}^{2} – y_{1}^{2}) + 2g (x_{2} – x_{1}) + 2f (y_{2} – y_{1}) = 0⇒ (x _{2} – x_{1}) (x_{1} + x_{2}) + (y_{2} – y_{1}) (y_{1} + y_{2}) + 2g (x_{2} – x_{1}) + 2f (y_{2} – y_{1}) = 0⇒ (y _{2} – y_{1}) / (x_{2} – x_{1}) = – (x_{1} + x_{2} + 2g) / (y_{1} + y_{2} + 2f) …………(iii)Now, equation to PQ is y – y _{1} = [ (y_{2} – y_{1}) / (x_{2} – x_{1}) ] (x – x_{1})⇒ y – y _{1} = [ – (x_{1} + x_{2} + 2g) / (y_{1} + y_{2} + 2f) ] (x – x_{1}) from (iii)As Q → P, x_{2} → x_{2} and y_{2} → y_{1} and PQ become tangent to the circle at P. The equation of the tangent then becomes.y – y _{1} = [ – { 2 (x_{1} + g) } / { 2 (y_{1} + f) } ] (x – x_{1})⇒ (y – y _{1}) (y_{1} + f) = – (x – x_{1}) (x_{1} + g)⇒ yy _{1} + fy – y_{1}^{2} – fy_{1} = – (xx_{1} + gx – x_{1}^{2} – gx_{1})⇒ xx _{1} + yy_{1} + gx + fy = x_{1}^{2} + y_{1}^{2} + gx_{1} + fy_{1}Adding (gx _{1} + fy_{1} + c) on both sides, we getxx _{1} + yy_{1} + g (x + x_{1}) + f (y + y_{1}) + c = x_{1}^{2} + y_{1}^{2} + 2gx_{1} + 2fy_{1} + c = 0Thus, the equation of the tangent to the circle x ^{2} + y^{2} + 2gx + 2fy + c = 0 at (x_{1}, y_{1}) isxx _{1} + yy_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0 The equation of the tangent to a curve at a point (xNote:_{1}, y_{1}) may be obtained from the equation of the curve replacing x^{2} by xx_{1}, y^{2} by yy_{1}, x by (1/2) (x + x_{1}), y by (1/2) (y + y_{1}) and retaining the constant terms. |

### Length of the Tangent from an External Point:

Let x^{2} + y^{2} + 2gx + 2fy + c = 0 be a circle and let P (x_{1}, y_{1}) be a point lying outside the circle. PT is tangent to the circle through P. Obviously, CT is the radius of the circle, and ∠CTP = 90°. The coordinates of C are (-g, -f) as shown in the figure.In the right triangle CTP, CP ^{2} = PT^{2} + CT^{2} ∴ PT ^{2} = CP^{2} – CT^{2} ⇒ PT ^{2} = (x_{1} + g)^{2} + (y_{1} + f)^{2} – [ √(g^{2} + f^{2} – c) ]^{2}⇒ PT ^{2} = x_{1}^{2} + 2gx_{1} + g^{2} + y_{1}^{2} + 2fy_{1} + f^{2} – g^{2} – f^{2} + c = 0⇒ PT ^{2} = x_{1}^{2} + y_{1}^{2} + 2gx_{1} + 2fy_{1} + c∴ PT =√( x _{1}^{2} + y_{1}^{2} + 2gx_{1} + 2fy_{1} + c) If S = 0 be a circle. Then, the length of the tangent drawn from an external point (xNote:_{1}, y_{1}) is given by √S_{1}, where S_{1} is the value of S obtained by putting x = x_{1}, y = y_{1}.PT ^{2} = x_{1}^{2} + y_{1}^{2} + 2gx_{1} + 2fy_{1} + cThe square of the length of the tangent drawn from an external point P (x _{1}, y_{1}) is called the power of point P with respect to the circle under consideration. |

Find the equation of tangent to the circle xExample 1:^{2} + y^{2} – 2x – 10y + 1 = 0 at (-3, -2). Equation of a tangent to the circle xSolution-^{2} + y^{2} – 2x – 10y + 1 = 0 at (-3, -2) is-x (-3) + y (-2) – 2 [ {x + (-3)} / 2 ] – 10 [ {y + (-2)} / 2 ] + 1 = 0 ⇒ -3x – 2y – x + 3 – 5y + 10 + 1 = 0 ⇒ -4x – 7y + 14 = 0 ⇒ 4x + 7y – 14 = 0 |

Find the equations of tangents to the circle xExample 2:^{2} + y^{2} – 6x + 4y – 12 = 0 which are parallel to the line 4x + 3y + 5 = 0. Let the equation of tangent parallel to the line 4x + 3y + 5 = 0 beSolution-4x + 3y + k = 0 ………..(i) Now, the equation of a circle is- x ^{2 }+ y^{2} – 6x + 4y – 12 = 0Here, the center (3, -2) and radius (r) = √[3 ^{2} + 2^{2 }– (-12) ]⇒ r = √(13 + 12) = √25 = 5 Since the line (i) is tangent to the circle, therefore the distance of the line (i) from the center (3, -2) of the circle is equal to the radius of the circle. ⇒ | 4(3) + 3(-2) + k | / √[ (4) ^{2} + (3)^{2 }] = 5⇒ | 6 + k | / 5 = 5 ⇒ | 6 + k | = 25 ⇒ 6 + k = ± 25 ⇒ 6 + k = 25 or 6 + k = -25 ⇒ k = 19 or k = -31 ∴ The required equation of tangent is, 4x + 3y + 19 = 0 and 4x + 3y – 31 = 0 |

The line 4x – 3y + 12 = 0 is a tangent at (-3, 0), and the line 3x + 4y = 16 is a tangent at (4, 1) to the circle. Find the equation of the circle.Example 3- The equation of the tangent at (-3, 0) to the circle is,Solution-4x – 3y + 12 = 0 ∴ The equation of Normal at (-3, 0) is- 3x + 4y + k = 0 ……….(i) Now (i) passes through (-3, 0) ∴ 3(-3) + 4(0) + k = 0 ⇒ k = 9 Equation (i) becomes- 3x + 4y + 9 = 0 ……….(a) Now equation of the tangent at (4, 1) to the circle is, 3x + 4y – 16 = 0 ∴ The equation of Normal at (4, 1) is- 4x – 3y + λ = 0 ……….(ii) Now (ii) passes through (4, 1) ∴ 4(4) – 3(1) + λ = 0 ⇒ 16 – 3 + λ = 0 ⇒ λ = -13 Equation (ii) becomes- 4x – 3y -13 = 0 ……….(b) Equations (a) and (b) represent the equation of the diameter of a circle which one solving gives the center of the circle. Solving (a) and (b), x / (-52 + 27) = y / (36 + 39) = 1 / (-9 – 16) ⇒ x/-25 = y/75 = 1/-25 ⇒ x = 1, y = -3 ∴ Coordinates of the center of the circle are (1, -3) Now, Radius = Distance between points (1, -3) and (-3, 0) ⇒ r = √ [ (-3 -1) ^{2} + (0 + 3)^{2}⇒ r = √(16 + 9) = √25 = 5 Thus, the equation of the circle is, (x – 1) ^{2} + [y – (-3)]^{2} = (5)^{2}⇒ x ^{2} + 1 – 2x + y^{2} + 9 + 6y = 25⇒ x ^{2} + y^{2} – 2x + 6y – 15 = 0 |

## Comments (No)