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The normal at the point (3, 4) on a circ...

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

`x^(2) +y^(2)+2x -2y - 13 = 0`

B

`x^(2) +y^(2) -2x -2y -11 = 0`

C

`x^(2)+y^(2)-2x +2y +12 = 0`

D

`x^(2) +y^(2) - 2x - 2y+14 = 0`

Text Solution

Verified by Experts

The correct Answer is:
B
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`
or `x^(2)+y^(2) -2x -2y -11 =0`
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