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

• A

  ${\left(x-9\right)}^2+{\left(y+8\right)}^2=25$
• Correct answer
  B

  ${\left(x-9\right)}^2+{\left(y-8\right)}^2=25$
• C

  ${\left(x+8\right)}^2+{\left(y+8\right)}^2=25$
• D

  ${\left(x+8\right)}^2+{\left(y-9\right)}^2=25$

Solution

The equation of the given circle is

$x^2+y^2-2x-4y-20=0$
Its centre is $C_1\left(\mathrm{1,2}\right)$ and radius $=5.$
The circle touches another circle of radius $5$ externally at $P\left(\mathrm{5,5}\right).$ Let its centre be $C_2\left(\alpha ,\beta \right).$
Clearly, $P\left(\mathrm{5,5}\right)$ is the midpoint of $C_1{C}_2.$ Therefore,
$\frac{\alpha +1}{2}=5$ and $\frac{\beta +2}{2}=5\Rightarrow \alpha =9,\beta =8.$
Hence, the equation of the required circle is
${\left(x-\alpha \right)}^2+{\left(y-\beta \right)}^2=5^2$
or, ${\left(x-9\right)}^2+{\left(y-8\right)}^2=5^2$

Hence, the correct answer is option: 'B'.