Question

Find the equation of the circle whose radius is 5 and which touches the circle
$x^{2}+$ $y^{2}-2 x-4 y-20=0$ 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

Hint: The location of a circle in the Cartesian plane is represented by a circle equation.
Here we have given the circle equation and the radius of the circle touches the given circle. So from the general formula of a circle we can find the center of the circles and we can form the required equation of the circle.

Formula used:
The equation of circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$

Complete Step by step solution:
The equation of the given circle is
${{x}^{2}}+{{y}^{2}}-2x-4y-20=0\text{ }$
Or ${{(x-1)}^{2}}+{{(y-2)}^{2}}={{5}^{2}}$
The equation for a circle has the generic form: ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$.
The coordinates of the circle's center and radius are found using this general form, where g, f, and c are constants.
From the circle equation
center is $C_{1}(1,2)$ and radius $=5$.
At point$P(5,5)$, this circle externally touches another circle of radius 5.
 let the center of the circle $C_{2}(\alpha, \beta)$.
Then the midpoint of$\mathrm{C}_{1} \mathrm{C}_{2}$ is $\mathrm{P}(5,5)$
Hence,
 $\dfrac{\alpha+1}{2}=5$ and $\dfrac{\beta+1}{2}=5$
$\Rightarrow \alpha=9, \beta=8$
Then
At point $(9,8)$, this circle externally touches another circle of radius 5.
Then the circle equation is
$(x-9)^{2}+(y-8)^{2}=5^{2}$.
Hence, the correct option is (B).

Note: In general,the radius of the circle, abbreviated $r$, is a constant that describes this fixed point, which is known as the circle's centre. The formula for a circle$\left(\mathrm{x}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}-\mathrm{y}_{1}\right)^{2}=\mathrm{r}^{2}$ whose centre is at $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$