 # Find the equation of a circle of radius `
Question:

Find the equation of a circle of radius 5 which is touching another circle x2 + y2 – 2x – 4y – 20 = 0 at (5, 5).

Solution:

Given x2 – 2x + y2 – 4y – 20 = 0

x2 – 2x + 1 +y2 – 4y + 4 – 20 – 5 = 0

(x – 1)2 + (y – 2)2 = 25

(x – 1)2 + (y – 2)2 = 52

Since, the equation of a circle having centre (h, k), having radius as r units, is

(x – h)2 + (y – k)2 = r2

Centre = (1, 2)

Point of Intersection = (5, 5)

It intersects the line into 1: 1, as the radius of both the circles is 5 units.

Using Ratio Formula,

$\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}$

Ratio $=m_{1}: m_{2}$

Assuming the co-ordinates of the centre of the circle be $(p, q)$

$5=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}$

$5=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}$

Now by substituting the values we get

$\frac{(1) p+(1) 1}{1+1}=\frac{p+1}{2}=5$

$\frac{1(q)+1(2)}{1+1}=\frac{q+2}{2}=5$

p + 1 = 10, q + 2 = 10

p = 9 & q = 8

Co-ordinates = (9, 8)

Therefore the equation is,

(x – h)2 + (y – k)2 = r2

(x – 9)2 + (y – 8)2 = 52

x2 – 18x + 81 + y2 – 16y + 64 = 25

x2 – 18x + y2 – 16y + 145 – 25 = 0

x2 – 18x + y2 – 16y + 120 = 0

Hence, the required equation is x2 – 18x + y2 – 16y + 120 = 0.