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  • The angle of intersection of the circles <img alt="\mathrm{x^2+y^2-6 x+4 y+11=0 \, \, and \, \, x^2+y^2-4 x+6 y+9=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus

The angle of intersection of the circles \mathrm{x^2+y^2-6 x+4 y+11=0 \, \, and \, \, x^2+y^2-4 x+6 y+9=0}, is

Option: 1

30^{\circ}


Option: 2

45^{\circ}


Option: 3

60^{\circ}


Option: 4

90^{\circ}


Answers (1)

best_answer

The equations of the circles are

\mathrm{ \begin{aligned} & x^2+y^2-6 x+4 y+11=0\, \, \, \, \, \, ...(i) \\ & x^2+y^2-4 x+6 y+9=0 \, \, \, \, \, ...(ii)\end{aligned} }   

The centres of these circles are \mathrm{C_1(3,-2) \, \, and \, \, C_2(2,-3)} respectively. Let r_1 and r_2 be their radii. Then,

\mathrm{ r_1=\sqrt{9+4-11}=\sqrt{2} \text { and } r_2=\sqrt{4+9-9}=2 \text {. } }

Also, C_1 C_2=\sqrt{2}

Suppose circles (i) and (ii) intersect at $P$. Then

\mathrm{ \begin{aligned} & \cos \angle C_1 P C_2=\frac{C_1 P^2+C_2 P^2-C_1 C_2^2}{2 C_1 P \cdot C_2 P}=\frac{2+4-2}{2 \times \sqrt{2} \times 2}=\frac{1}{\sqrt{2}} \\ & \Rightarrow \angle C_1 P C_2=45^{\circ} \end{aligned} }

Thus, the angle of intersection of the given circles is of \mathrm{45^{\circ}.}

Posted by

jitender.kumar

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