Question

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

• A. $ (x-9)^2 + ( y-8)^2 = 5^2 $
• B. $ (x-5)^2 + ( y-5)^2 = 5^2 $
• C. $ (x-0)^2+(y-0)^2=5^2 $
• D. None of the above

Answer 1

Answer: (A)

Let equation of circle be $(x-h)^{2}+(y-k)^{2}=5^{2}$,
where $5$ is the radius of the circle.
Also, the radius of given circle
$c_{1}: x^{2}+y^{2}-2 x-4 y-20=0$ is given by
$\sqrt{1+4+20}=5$ units

$\because$ Two circles touch each other externally.
$\therefore(5,5)$ is the mid-point of line segment joining $(h, k)$ and $(1,2)$
$\Rightarrow 5=\frac{h+1}{2}, 5=\frac{k+2}{2}$
$\Rightarrow h=9, k=8$
$\therefore$ Equation of circle becomes
$(x-9)^{2}+(y-8)^{2}=5^{2}$