Question

Guys please help !!!
The equation of the circle concentric with the circle x²+y²+8x+10y-7 = 0 and passing through the centre of the circle x²+y²-4x-6y = 0​

Answer 1

Question :

The equation of the circle concentric with the circle x²+y²+8x+10y-7 = 0 and passing through the centre of the circle x²+y²-4x-6y = 0

Theory :

•Genral equation of a circle :

The general equation of a circle is $\sf\:x{}^{2}+y{}^{2}+2gx+2fy+c=0$, where g,f,c are constants .

Centre of circle (-g,-f)

i.e $\sf\:Centre=(\frac{-1}{2}\:cofficeint\:of\:x,\frac{-1}{2}\:cofficeint\:of\:y)$

Radius of the circle is $\sf\:\sqrt{g{}^{2}+f{}^{2}-c}$

•Central form of equation of a circle

The equation of circle having cente (h,k) and radius r is

(x-h)²+(y-k)²=r²

Solution :

Given : $\sf\:C_{1}\:x{}^{2}+y{}^{2}+8x+10y-7=0$

$\sf\:C_{2}\:x{}^{2}+y{}^{2}-4x-6y=0$

We have to find the equation of $\sf\:C_{3}$

___________________________

Centre of $\sf\:C_{1}=(\frac{-8}{2},\frac{-10}{2})=(-4,-5)$

Centre of $\sf\:C_{2}=(\frac{-(-4)}{2},\frac{-(-6)}{2})=(2,3)$

According to the question,

$\sf\:C_{3}$ pases through the centre of $\sf\:C_{2}$

and concentric with the circle $\sf\:C_{1}$

Therefore,

Centre of $\sf\:C_{3}=(-4,-5)$

and $\sf\:C_{3}$ passes through the point (2,3).

Let (4,-5) be A and (2,3) be P

By using distance formula :

Radius of circle $\sf\:C_{3}$

= length of AP

$= \sqrt{(2 - ( - 4)) {}^{2} + (3 - ( - 5)) {}^{2} }$

$= \sqrt{6 {}^{2} + 8 {}^{2} } = \sqrt{100} = 10$

Now by using central form of equation of a circle .

The equation of circle $\sf\:C_{3}$ is $\sf(x - 2) {}^{2} + (y - 3) {}^{2} = 100$