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Question

If h2+k2>a2 , then the number of solutions of the equation hx+ky=a2 and x2+y2=a2 is

A
1
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B
2
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C
4
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D
0
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Solution

The correct option is B 2
x2+y2=a2
Given h2+k2>a2
then p(h,k) lies outside circle x2+y2=a2
Then chord of contact from p(h,k) to x2+y2=a2 is
xh+yk=a2
So, xh+yk=a2
no. of solution of the equation hx+ky=a2 and x2+y2=a2 are 2.

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