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
$c_1:x^2+y^2-2x-4y-20=0$ 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
$(x-9{)}^2+(y-8{)}^2=5^2$