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

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 (A) x2+y2+2 x-2 y-13=0 (B) x2+y2-2 x-2 y-11=0 (C) x2+y2-2 x+2 y+12=0 (D) x2+y2-2 x-2 y+14=0

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/10f7ede80060dca8dda0f524271b46c4-.png" /> (x-3)(x+1)+(y-4)(y+2)=0 Equation x2+y2-2 x-2 y-11=0

Q. 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

$x^{2} + y^{2} + 2 x - 2 y - 13 = 0$

$x^{2} + y^{2} - 2 x - 2 y - 11 = 0$

$x^{2} + y^{2} - 2 x + 2 y + 12 = 0$

$x^{2} + y^{2} - 2 x - 2 y + 14 = 0$

$(x - 3) (x + 1) + (y - 4) (y + 2) = 0$
Equation $x^{2} + y^{2} - 2 x - 2 y - 11 = 0$

Q. 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

x2+y2+2x−2y−13=0

x2+y2−2x−2y−11=0

x2+y2−2x+2y+12=0

x2+y2−2x−2y+14=0