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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Mathematics
Chapter
Circle

BITSAT 2014 · Mathematics · Circle

The angle of intersection of the two circles x^ 2 +y^ 2 -2 x-2 y=0...

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

Options

  1. A30∘30^{{}^{\circ}}
  2. B60∘60^{{}^{\circ}}
  3. C90∘90^{{}^{\circ}}
  4. D45∘45^{{}^{\circ}}

Correct answer

D. 45∘45^{{}^{\circ}}

Step-by-step solution

Here circles are x2+y2−2x−2y=0...(1)x2+y2=4...(2) Now, C1(1,1),r1=12+12=2C2(0,0),r2=2\begin{array}{l} x^{2}+y^{2}-2 x-2 y=0...(1) \\ x^{2}+y^{2}=4...(2) \\ \text{ Now, } \mathrm{C}_{1}(1,1), r_{1}=\sqrt{1^{2}+1^{2}}=\sqrt{2}\\ C_{2}(0,0), r_{2}=2 \end{array} If θ\theta is the angle of intersection then cos⁡θ=r12+r22−(c1c2)22r1r2=2+4−(2)22⋅2.2=12⇒θ=45∘\begin{array}{l} \cos \theta=\frac{r_{1}^{2}+r_{2}^{2}-\left(c_{1} c_{2}\right)^{2}}{2 r_{1} r_{2}}\\ =\frac{2+4-(\sqrt{2})^{2}}{2 \cdot \sqrt{2.2}}=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{{}^{\circ}} \end{array}
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