Question

the normal at the point (3,4) on a circle cuts the circle at the point (-1,-2).then the equation of the circle is

Answer 1

Let C(h,k) be the center of the circle.

A(3,4) and B(-1,-2) and these both points  lie on circle.

Mid point of AB will be centre C (h,k)

C (h,k) = (3-1/2, 4-2/2)

= (1,1)

AC = radius of circle = r = √(3-1)² + (1-4)²

= √4 + 9

= √13

Now the equation of circle will be:

(x-h)² + (y-k)²= r²

(x-1)² + (y-1)² = (√13)²

= (x-1)² + (y-1)²

= 13

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