Tangent Circle Radius

If a point moves with respect to a fixed point by maintaining a constant distance between itself and the fixed point, it is called a circle. The fixed point is called the center of the circle and the constant distance is called the radius of the circle.

If (x1, y1) is the center and r is the radius of the circle, the equation of the circle is given by (x – x1)2 + (y – y1)2 = r2 ………….. (1)

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If the center is the origin, (1)  `=>` x2 + y 2 = r 2 ………. (2)

The general equation of the circle is given by x 2 + y 2 + 2gx + 2fy + c = 0 …….. (3)

[(1) Can be simplified to (3)]

Its center is (-g, -f) and radius is `sqrt [g ^2 + f ^2 ** c]` .

If a line touches the circle at the point (x1, y1) it is called the tangent and it is given by as follows: From (2) we have the tangent equation as xx1 + yy1 = r2 ………… (T1)

From (3), we have the tangent equation as

xx1 + yy1+g( x + x1) + f (y + y1) + c = 0 ……….. (T2)
Example Problems on Tangent Circle Radius.

Ex 1: If (2, 3) and 5 are the center and radius of the circle, find its equation.

Soln: Given: (x1, y1) = (2, 3) and r = 5

Therefore By using (1), we get: (x – x1) 2 + (y – y1) 2 = r 2

`=>` (x – 2) 2 + (y -3)2 = 52

`=>` x 2 – 4x + 4 + y 2 – 6y + 9 = 25

`=>` x 2 + y 2 – 4x – 6y – 12 = 0.

Hence the equation of the circle.

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Ex 2: Find the center, radius and the equation of tangent of the circle

x 2 + y 2 – 10x + 4y – 140 = 0 at (-7, -7) on it.

Soln: Given: x 2 + y 2 – 10x + 4y – 140 = 0 …………. (1)

From (1), 2g = -10 `=>` g = -5

2f = 4 `=>` f = 2

Therefore the center is (-g, -f) = (5, -2)

In (1), C = -140

Therefore Radius = `sqrt [g ^2 + f ^2 ** c]` = `sqrt [5 ^2 + (-2) ^2 ** (-140)]`

`=>` `sqrt [25 + 4 + 140]` = `sqrt 169` = 13.

Therefore Radius = 13.

The equation of the tangent at (-7, -7), we get:

 xx1 + yy1 – 5 (x + x1) + 2 (y + y1) – 140 = 0

`=>` -7x – 7y – 5x + 35 + 2y – 14 -140 = 0

`=>` -12 x – 5y – 119 = 0 = 12x + 5y + 119 = 0.
More Example Problems on Tangent Circle Radius.

Ex 3: Find the equation of the tangent at (-2, 3) to the circle x2 + y 2 – 13 = 0.

Soln: Given: (x1, y1) = (-2, 3).

The tangent is x (-2) + y (3) – 13 = 0

`=>` - 2x + 3y – 13 = 0

`=>` 3y = 2x + 13.