Question

Find the equation of the circle cocentric with the circle x^2+y^2+4x-6y-13=0 and passes through the centre of the circle x^2+y^2-8x-10y-8=0?

Answer 1

Let's,find the centre of the circle whose equation is to be made,

given, `x^2 + y^2 +4x -6y-13=0`

or, `(x+2)^2 +(y-3)^2 =26`

So,its centre will be at `(-2,3)`

Now,centre of the other circle can be found in the similar way,

`x^2 +y^2-8x-10y-8=0`

or, `(x-4)^2 + (y-5)^2 =49`

So,our circle of concerned will be passing through `(4,5)`

So,if the equation of the new circle be `(x+2)^2 +(y-3)^2 =r^2` (where, `r` is the radius of the circle)

As it passes through `(4,5)`,we can write,

`(4+2)^2 +(5-3)^2 = r^2`

or, `r^2 = 40`

So,the equation of the new circle is `(x+2)^2 +(y-3)^2 = 40`