Question

Find the equation of the circle whose centre lies at point (2,2) and passes through the centre of the circle x^2+y^2+4x+6y+2=0

Answer 1

hope this will help. ...

Answer 2

It is given that the circle is centered at ($2$, $2$) and assuming the radius $r^{}$ , then the equation of the circle is

$(x-2)^{2}$ + $(y - 2)^{2}$ = $r^{2}$,

Now this circle passes through the circle -

$x^{2}  + y^{2} + 4x + 6y + 2 = 0$  OR

$(x + 2)^{2} + (y + 3)^{2} - 4 - 9 + 2 = 0$ OR

$(x + 2)^{2} + (y + 3)^{2} = 11$

whose center is $(-2, -3)$

Therefore the circle whose equation we found out passes through (-2, -3) and therefore this point will satisfy that equation.

$4^{2} + 6^{2} = r^{2}$ OR $40 = r^{2}$

The required equation is,

$(x - 2)^{2} + (y - 2)^{2} = 40$