The normal at the point (3, 4) on a circle cuts the circle at the point (-1, -2). Then…
Mathematics · JEE Main · NTA Exams — Co-ordinate Geometry
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,
- \(x^{2}+y^{2}+2 x-2 y-13=0\)
- \(x^{2}+y^{2}-2 x-2 y-11=0\)
- \(x^{2}+y^{2}-z x+z y+12=0\)
- \(x^{2}+y^{2}-2 x-2 y+14=0\)
Answer
(B) x^ 2 +y^ 2 -2 x-2 y-11=0
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