IIT JEE 1983 Maths - MCQ Question 25

Geometry Level 2

The equation of the circle passing through the point ( 1 , 1 ) (1, 1) and the points of intersection of x 2 + y 2 + 13 x − 3 y = 0 x^2+y^2+13x-3y=0 and 2 x 2 + 2 y 2 + 4 x − 7 y − 25 = 0 2x^2+2y^2+4x-7y-25=0 is ___ . \text{\_\_\_}.

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4 x 2 + 4 y 2 − 17 x − 10 y + 25 = 0 4x^2+4y^2-17x-10y+25=0 4 x 2 + 4 y 2 − 30 x − 10 y − 25 = 0 4x^2+4y^2-30x-10y-25=0 4 x 2 + 4 y 2 + 30 x − 13 y − 25 = 0 4x^2+4y^2+30x-13y-25=0 None of the others

Shubhamkar Ayare by Shubhamkar Ayare , 22, India

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1 solution

Rishabh Jain
Jan 7, 2017

The circle passing through intersection is :

x 2 + y 2 + 13 x − 3 y + λ ( 2 x 2 + 2 y 2 + 4 x − 7 y − 25 ) = 0 x^2+y^2+13x-3y+\color{#D61F06}{\lambda}\color{#333333}{(2x^2+2y^2+4x-7y-25)=0}

Since it passes through ( 1 , 1 ) (1,1) , putting we get: 12 + λ ( − 24 ) = 0 ⟹ λ = 1 2 12+\lambda(-24)=0\implies \color{#D61F06}{\lambda}=\dfrac{1}{2}

Thus required circle is: x 2 + y 2 + 13 x − 3 y + 1 2 ( 2 x 2 + 2 y 2 + 4 x − 7 y − 25 ) = 0 x^2+y^2+13x-3y+\color{#D61F06}{\dfrac 12}\color{#333333}{(2x^2+2y^2+4x-7y-25)=0} ⟹ 4 x 2 + 4 y 2 + 30 x − 13 y − 25 = 0 \implies 4x^2+4y^2+30x-13y-25=0

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Does the question in the problem asks for the equation of the circle that passes through (1,1) and the points of intersection of the two given circles? Because i plotted the two circles then located the points of intersection which is (-1.4,5.8) and (-1.031,-2.32). So the equation of the circle must pass through (1,1) , (-1.4,5.8) and (-1.031,-2.32).

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