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10-1.Circle and System of Circles
medium
The point at which the normal to the circle ${x^2} + {y^2} + 4x + 6y - 39 = 0$ at the point $(2, 3)$ will meet the circle again, is
A
$(6, -9)$
B
$(6, 9)$
C
$(-6, -9)$
D
$(-6, 9)$
Solution
(c) Equation of normal will be $\frac{{x – 2}}{{2 + 2}} = \frac{{y – 3}}{{3 + 3}}$
$ \Rightarrow $$3x – 2y = 0$
$\Rightarrow x = \frac{{2y}}{3}$
Thus ${\left( {\frac{{2y}}{3}} \right)^2} + {y^2} + 4\left( {\frac{{2y}}{3}} \right) + 6y – 39 = 0$
$ \Rightarrow y = 3,\; – 9 $
$\Rightarrow x = 2,\; – 6$
Hence another point will be $( – 6,\, – 9)$.
Standard 11
Mathematics
