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10-1.Circle and System of Circles
normal
If variable point $(x, y)$ satisfies the equation $x^2 + y^2 -8x -6y + 9 = 0$ , then range of $\frac{y}{x}$ is
A
$\left[ { - \frac{7}{{24}},\frac{7}{{24}}} \right]$
B
$\left[ { - \frac{7}{{24}},\infty } \right)$
C
$\left[ {\frac{7}{{24}},\infty } \right)$
D
$\left( { - \infty ,\infty } \right)$
Solution
Let $\mathrm{y}=\mathrm{mx}$ is a tangent
to the circle
$x^{2}+y^{2}-8 x-6 y+9=0$
then length of perpendicular $=$ Radius
$\frac{|4 m-3|}{\sqrt{1+m^{2}}}=4 \Rightarrow(4 m-3)^{2}=16\left(1+m^{2}\right)$
$\Rightarrow 16 \mathrm{m}^{2}+9-24 \mathrm{m}=16+16 \mathrm{m}^{2}$
$\Rightarrow \quad 24 \mathrm{m}=-7$
$m=-\frac{7}{24}$
Standard 11
Mathematics
