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  • Find the equation of the circle through the points of intersection of the circles <img alt="\mathrm{x^2+y^2-4 x-6 y-12=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7

Find the equation of the circle through the points of intersection of the circles \mathrm{x^2+y^2-4 x-6 y-12=0} and \mathrm{x^2+y^2+6 x+4 y-12=0} and intersecting the circle \mathrm{x^2+y^2-2 x-4=0} orthogonally.



 

Option: 1

\mathrm{x^2+y^2+8 x+7 y-12=0}


Option: 2

\mathrm{x^2+y^2+16 x+14 y-6=0}


Option: 3

\mathrm{x^2+y^2+8 x+7 y-6=0}


Option: 4

\mathrm{x^2+y^2+16 x+14 y-12=0}


Answers (1)

best_answer

The equation of the circle through the points of intersection of the given circles is


\mathrm{x^2+y^2-4 x-6 y-12+\lambda(-10 x-10 y)=0} -----------------(1)

Equation (1) can be rearranged as

\mathrm{\mathrm{x}^2+\mathrm{y}^2-\mathrm{x}(10 \lambda+4)-\mathrm{y}(10 \lambda+6)-12=0 \text {. Its centre is }(5 \lambda+2,5 \lambda+3)}

The centre of the circle \mathrm{\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{x}-4=0 \text { is }(1,0)}

The condition of orthogonality is \mathrm{2 \mathrm{gg}_1+2 \mathrm{ff}_1=\mathrm{c}+\mathrm{c}_1}

\mathrm{\text { Hence } 2(5 \lambda+2)(1)+2(5 \lambda+3)(0)=-12-4 \Rightarrow \lambda=-2}4

Hence the required circle is


\mathrm{x^2+y^2-4 x-6 y-12-2(-10 x-10 y)=0 \quad \text { i.e., } x^2+y^2+16 x+14 y-12=0}

 

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