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  • The normal at the point (3,4) on a circle cuts the circle at the point (-1,-2). Then the equation of the circle isOption: 1 <img alt="\mathrm{x^

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

Option: 1

\mathrm{x^2+y^2+2 x-2 y-13=0}


Option: 2

\mathrm{x^2+y^2-2 x-2 y-11=0}


Option: 3

\mathrm{x^2+y^2-2 x+2 y+12=0}


Option: 4

\mathrm{x^2+y^2-2 x-2 y+14=0}


Answers (1)

best_answer

Clearly, (3,4) and (-1,-2) are end points of diameter.

Thus, equation of circle is (x-3)(x+1)+(y-4)(y+2)=0

\Rightarrow x^2+y^2-2 x-2 y-11=0

Posted by

Divya Prakash Singh

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