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

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

If (-10,2) is an interior point, the circle is more likely to touch x-y=0 at P(-4,-4) in the 3^{\text {rd }} quadrant

since O P=4 \sqrt{2}.
The line x+y=0 meets the circle at A and B such that A B=6 \sqrt{2} ; \quad C M=O P=4 \sqrt{2} and

A M=\frac{1}{2} A B=3 \sqrt{2}.

C A=\sqrt{C M^2+A M^2}=5 \sqrt{2}= radius of the circle =C P.

\therefore C  lies on P C whose equation is x+y+8=0 and on C M whose equation is x-y+10=0.

This is for the reason that if M be (-x, x) then \sqrt{x^2+x^2}=5 \sqrt{2} giving x=5 \, i.e, M is

(-5,5). x+y+8=0 and x-y+10=0 when solved simultaneously fixes the centre at (-9,1).

\therefore  the equation to the circle is (x+9)^2+(y-1)^2=(5 \sqrt{2})^2

i.e., x^2+y^2+18 x-2 y+32=0                           ...(i)

Substituting x=-10, y=2 on the L.H.S. of the above equation, we get 100+4-180-4+32=-48<0.

\therefore(-10,2) lies inside the circle.

Hence the circle (i) is the required circle since it satisfies all the conditions stated in the problem.

 

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