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10-1.Circle and System of Circles
medium
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
${x^2} + {y^2} - 2x - 2y + 14 = 0$
Solution
(b) Since normal passes through the centre of the circle.
$\therefore $ The required circle is the circle with ends of diameter as $(3, 4)$ and $(-1, -2).$
$\therefore $ It's equation is $(x – 3)\,\,(x + 1)\, + (y – 4)(y + 2) = 0$
$ \Rightarrow $ ${x^2} + {y^2} – 2x – 2y – 11 = 0.$
Standard 11
Mathematics