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Tardigrade
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Mathematics
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
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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
Conic Sections
A
$x^2+y^2+2 x-2 y-13=0$
B
$x^2+y^2-2 x-2 y-11=0$
C
$x^2+y^2-2 x+2 y+12=0$
D
$x^2+y^2-2 x-2 y+14=0$
Solution:
$(x-3)(x+1)+(y-4)(y+2)=0$
Equation $x^2+y^2-2 x-2 y-11=0$
