Find the equation of the circle concentric with the circle

Question:

Find the equation of the circle concentric with the circle $x^{2}+y^{2}-6 x+12 y+$ 15 = 0 and of double its area.

 

Solution:

2 or more circles are said to be concentric if they have the same centre and different radii.

Given, $x^{2}+y^{2}-6 x+12 y+15=0$

Radius r =

$\sqrt{g^{2}+f^{2}-c}=\sqrt{\left(-3^{2}\right)+6^{2}-15}=\sqrt{30}$

The concentric circle will have the equation

$x^{2}+y^{2}-6 x+12 y+c^{\prime}=0$

Also given area of circle $=2 \times$ area of the given circle.

$\Rightarrow r^{\prime 2}=2 \times r^{2}=2 \times 30=60$

We can get $c^{\prime}=45-60=-15$

The required equation is $x^{2}+y^{2}-6 x+12 y-15=0$.

 

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