Question

find the equation of the circle passing through the point (-3,2) and centre of the circle is (2 ,4)​

Answer 1

radius is

$\sqrt{ {5}^{2} + {2}^{2} } = \sqrt{29}$

general form of circle equation....

$( {x - 2})^{2} + ({y - 4})^{2} = 29$

Answer 2

ATQ,

x=-3,y=2,h=2,k=4

Now,

By standard equation of circle.

(x-h)^2 + (y-k)^2 = r^2

=> r^2 = (-3-2)^2 + (2-4)^2

=> r^2 = (-5)^2 + (-2)^2

=> r^2 = 25 + 4

=> r^2 =29

Now,

By standard equation of circle.

(x-h)^2 +(y-k)^2 = r^2

=> (x-2)^2 + (y-4)^2 =29

=> x^2 + 4 -4x + y^2 +16 -8y -29 = 0

=> x^2 + y^2 -4x -8y - 9 =0

****HOPE YOU UNDERSTAND ****

THANK YOU