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10-1.Circle and System of Circles
medium
વર્તૂળો $x^2 + y^2 + 8x - 2y - 9 = 0$ અને $x^2+ y^2 -2x + 8y - 7 = 0$ નો છેદ કોણ : ............ $^o$
A
$60$
B
$90$
C
$45$
D
$30$
Solution
$x ^2+ y ^2+8 x -2 y -9=0$
$C _1=(-4,1) \text { and } r _1=\sqrt{4^2+1^2+9}=\sqrt{26}$
$x ^2+ y ^2-2 x +8 y -7=0$
$C _2 \equiv(1,-4) \text { and } r _2=\sqrt{1^2+4^2+7}=\sqrt{24}$
$\therefore \cos (180-\theta)=\frac{ r _1{ }^2+ r _2{ }^2-\left( C _1 c _2\right)^2}{2 r _1 r _2}$
$\Rightarrow-\cos \theta=\frac{24+26-50}{2 \sqrt{24} \sqrt{26}}$
$\Rightarrow \cos \theta=0$
$\Rightarrow \theta=90^{\circ}$ is the angle between the circles
Standard 11
Mathematics
