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Question

Mathematics Question on Circle

The angle of intersection of the two circles x2+y22x2y=0x^2 + y^2 - 2x - 2y = 0 and x2+y2=4x^2 + y^2 = 4, is

A

3030^{\circ}

B

6060^{\circ}

C

9090^{\circ}

D

4545^{\circ}

Answer

4545^{\circ}

Explanation

Solution

Here circles are
x2+y22x2y=0x^2 + y^2 - 2x - 2y = 0 ...(1)
x2+y2=4x^2 + y^2 = 4 ...(2)
Now, C1(1,1),r1=12+12=2C_1 (1, 1), r_1 = \sqrt{1^2 + 1^2} = \sqrt{2}
C2(0,0),r2=2C_2 (0,0) , r_2 = 2
If θ\theta is the angle of intersection then
cosθ=r12+r22(c1c2)22r1r2\cos\theta = \frac{r^{2}_{1} + r^{2}_{2} -\left(c_{1}c_{2}\right)^{2}}{2r_{1}r_{2}}
=2+4(2)222.2.=12= \frac{2+4 -\left(\sqrt{2}\right)^{2}}{2\sqrt{2}.2.}= \frac{1}{\sqrt{2}}
θ=45\Rightarrow \theta = 45^{\circ}