Question

Find the equation of circle whose radius is 5 and which touches the circle $x^2+y^2-2x-4y-20=0$ at the point (5, 5).

• A. x² + y² – 18x – 16y + 120 =0
• B. x² + y² + 6x – 3y - 54 =0
• C. x² + y² – 17x – 19y + 50 =0
• D. x² + y² – 18x – 8y + 60 =0

Answer 1

The correct option is A x² + y² – 18x – 16y + 120 =0

The given circle is $x^2+y^2-2x-4y-20=0$ with centre (1, 2) and radius = $\sqrt{1+4+20}=5$.

And required circle has radius 5 hence circles touch each other externally.

Since point of contact is P(5, 5).

P is the mid point of $C_1$ and $C_2$, let co – ordinate of centre $C_2$ is (h, k) then

$\frac{1+h}{2}=5;h=9$

$\frac{2+k}{2}=5$; k = 8

And, Equation of required circle is $(x-9{)}^2+(y-8{)}^2=5^2$

$x^2+y^2-18x-16y+120=0;$ Option(a)