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Question

Find the equation to that chord of the circle x2+y2=81 which is bisected at the point (2,3), and its pole with respect to the circle.

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Solution

Let the chord be y=mx+c ... (i)

The line joining the midpoint of the chord to centre of the circle is

perpendicular to the circle, whose slope will be 1m

1m=3020

m=23

To calculate c, put value (2,3) in the equation (i) of line

323×(2)=c

c=133

Thus, the chord is given by 3y2x=13

Let the pole of this chord be (h,k)

Polar of (h,k) is given by xh+yk=81

Since the chord and polar are same line, there equation must be identical
h2=k3=8113

(h,k)=(16213,24313)

692397_640973_ans_9539ff9781a64e17a09c88c72ee1310c.png

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