Q. 265.0( 2 Votes )

# Find the equation of a circle of radius 5 which is touching another circlex2 + y2 – 2x – 4y – 20 = 0 at (5, 5).

Answer :

x^{2} - 2x + y^{2} - 4y – 20 = 0

x^{2} - 2x + 1 +y^{2} - 4y +4 – 20 – 5 = 0

(x – 1)^{2} + (y – 2)^{2} = 25

(x – 1)^{2} + (y – 2)^{2} = 5^{2}

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

(x – h)^{2} + (y – k)^{2} = r^{2}

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,

Ratio = m_{1} : m_{2}

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

p + 1 = 10, q + 2 = 10

p = 9 & q = 8

Co-ordinates = (9,8)

Equation is,

(x – h)^{2} + (y – k)^{2} = r^{2}

(x – 9)^{2} + (y – 8)^{2} = 5^{2}

x^{2} - 18x + 81 + y^{2} - 16y + 64 = 25

x^{2} - 18x + y^{2} - 16y + 145 – 25 = 0

x^{2} - 18x + y^{2} - 16y + 120 = 0

Hence, the required equation is x^{2} - 18x + y^{2} - 16y + 120 = 0.

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