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The extremities of a diagonal of a rectangle are(-4,4) and (6,-1). A circle circumscribes the rectangle and cuts an intercept \mathrm{AB} on the \mathrm{y}-axis. Find the point of intersection of the tangents to the circle at A and B.

Option: 1

\left(\frac{121}{4}, \frac{3}{2}\right)


Option: 2

\left(\frac{-121}{4}, \frac{1}{4}\right)


Option: 3

\left(\frac{-121}{4}, \frac{3}{2}\right)


Option: 4

\left(\frac{121}{4}, \frac{-3}{4}\right)


Answers (1)

best_answer

So, the equation of the circle is (x+4)(x-6)+(y-4)(y+1)=0
Or, x^2+y^2-2 x-3 y-28=0
Let the required point is (h, k)
The chord of contact w.r.t (h, k) is y-axis i.e. x=0   ..........(1)
It is also T=0
i.e. x h+y k-(x+h)-\frac{3}{2}(y+k)-28=0
i.e. x + \frac{\left(k-\frac{3}{2}\right) y}{h-1}-\frac{\left(h+\frac{3}{2} k+28\right)}{h-1}=0    ................(2)
comparing (1) and (2), we get k=\frac{3}{2} and h=-\frac{121}{4}
so, the point is
\left(\frac{-121}{4}, \frac{3}{2}\right)

Posted by

Anam Khan

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