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  • A circle touches the line y=x at a point P such that , where O i

A circle touches the line y=x at a point P such that \mathrm{O P=4 \sqrt{2}}, where O is the origin. The circle contains the point (-10,2) in its interior and the length of its chord on the line x+y=0 is \mathrm{6 \sqrt{2}}. The equation of the circle is

Option: 1

\mathrm{x^2+y^2+18 x-2 y+32=0}


Option: 2

\mathrm{x^2+y^2+36 x-4 y+64=0}


Option: 3

\mathrm{x^2+y^2+9 x-y+16=0}


Option: 4

\mathrm{x^2+y^2+4 x+2 y+16=0}


Answers (1)

best_answer

: Let CL be the perpendicular to the chord QR. OPCL is a rectangle in which \mathrm{O P=4 \sqrt{2}=C L}

\mathrm{\begin{aligned} & Q L=\frac{1}{2} Q R=3 \sqrt{2} \\ & \therefore \quad r=C Q=\sqrt{C L^2+L Q^2}=\sqrt{(4 \sqrt{2})^2+(3 \sqrt{2})^2}=5 \sqrt{2} \\ & \therefore \quad C P=5 \sqrt{2} \end{aligned}}

\mathrm{\text { Let } C \text { be }(h, k) \text {. Then }\left|\frac{k-h}{\sqrt{2}}\right|=5 \sqrt{2}}

\mathrm{\therefore k-h= \pm 10}     ...[1]

\mathrm{\text { Also, }\left|\frac{k+h}{\sqrt{2}}\right|=4 \sqrt{2} \Rightarrow k+h= \pm 8}      ....[2]

(1) and (2) will make four sets of equations which when solved will give the centres as
(9, –1), (1, –9), (–1, 9), (–9, 1)

Among these (–9, 1) is the only centre which will make (–10, 0) an interior point of the circle

\mathrm{\therefore \quad \text { Equation of the circle is }(x+9)^2+(y-1)^2=(5 \sqrt{2})^2}

\mathrm{\text { or, } x^2+y^2+18 x-2 y+32=0}

Posted by

himanshu.meshram

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