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

answer

x^2+y^2-x-2y-14=0

solution

(x-x1)(x-x2)+(y-y1)(y-y2)=0

(x1,y1)=(3,4)

(x2,y2)=(-1,-2)

(x-3)(x-(-2))+(y-4)(y-(-2))=0

(x-3)(x+2)+(y-4)(y+2)=0

x^2-3x+2x-6+y^2-4y+2y-8=0

x^2+y^2-x-2y-14=0

x^2+y^2+2fx+2gy+d=0