Mathematics - Circle Question with Solution | TestHub

MathematicsCircleRadical Axis & Radical Centre, Coaxial system of circles, Orthogonality of two circlesMedium2 minPYQ_2014

MathematicsMediumsingle choice

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

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Here circles are $x^{2} + y^{2} - 2 x - 2 y = 0... (1) x^{2} + y^{2} = 4... (2) Now, C_{1} (1, 1), r_{1} = 1^{2} + 1^{2} = 2 C_{2} (0, 0), r_{2} = 2$ If $θ$ is the angle of intersection then $cos θ = \frac{r _{1}^{2} + r _{2}^{2} - ( c _{1} c _{2} ) ^{2}}{2 r _{1} r _{2}} = \frac{2 + 4 - ( 2 ) ^{2}}{2 \cdot 2.2} = \frac{1}{2} \Rightarrow θ = 4 5^{\circ}$

Stream:BITSATSubject:MathematicsTopic:CircleSubtopic:Radical Axis & Radical Centre, Coaxial system of circles, Orthogonality of two circles

⏱ 2mℹ️ Source: PYQ_2014

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