Question

Find 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 $(5,5)$.

• A. $(x-8{)}^2+(y-9{)}^2=5^2$
• B. $(x-9{)}^2+(y-8{)}^2=5^2$
• C. $(x-8{)}^2+(y-8{)}^2=5^2$
• D. none of these

Answer 1

The correct option is B $(x-9{)}^2+(y-8{)}^2=5^2$

The equation of the given circle is

$x^2+y^2-2x-4y-20=0$ or, $(x-1{)}^2+(y-2{)}^2=5^2$.

Its centre is $C_1(1,2)$ and radius = 5. This circle touches another circle of radius 5 externally at point P(5,5). let its centre be $C_2(\alpha ,\beta )$.

Clearly, $P(5,5)$ is the mid-point of $C_1{C}_2$.

Therefore, $\frac{\alpha +1}{2}=5$ and $\frac{\beta +1}{2}=5$

$\Rightarrow \alpha =9,\beta =8$

Thus, the required circle has its centre at $(9,8)$ and radius $=5.$

Hence, the equation of the required circle is $(x-9{)}^2+(y-8{)}^2=5^2$.