If a tangent to the circle x^2 + y^2 = 1 inte | vedclass

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(C) Let the mid-point of $PQ$ be $S(h, k).$Since $P$ lies on the $x$-axis and $Q$ lies on the $y$-axis,let $P = (a, 0)$ and $Q = (0, b).$The mid-point

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$x^2 + y^2 - 4x^2y^2 = 0$

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(C) Let the mid-point of $PQ$ be $S(h, k).$Since $P$ lies on the $x$-axis and $Q$ lies on the $y$-axis,let $P = (a, 0)$ and $Q = (0, b).$The mid-point $S(h, k)$ is given by $h = \frac{a}{2}$ and $k = \frac{b}{2},$ so $a = 2h$ and $b = 2k.$The equation of the line $PQ$ is $\frac{x}{a} + \frac{y}{b} = 1,$ which becomes $\frac{x}{2h} + \frac{y}{2k} = 1.$Since this line is a tangent to the circle $x^2 + y^2 = 1,$ the perpendicular distance from the origin $(0, 0)$ to the line must be equal to the radius $r = 1.$The distance $d = \frac{|-1|}{\sqrt{(\frac{1}{2h})^2 + (\frac{1}{2k})^2}} = 1.$Squaring both sides,we get $\frac{1}{\frac{1}{4h^2} + \frac{1}{4k^2}} = 1,$which simplifies to $\frac{1}{4h^2} + \frac{1}{4k^2} = 1.$Multiplying by $4h^2k^2,$ we get $k^2 + h^2 = 4h^2k^2.$Replacing $(h, k)$ with $(x, y),$ the locus is $x^2 + y^2 = 4x^2y^2$ or $x^2 + y^2 - 4x^2y^2 = 0.$

If a tangent to the circle $x^2 + y^2 = 1$ intersects the coordinate axes at distinct points $P$ and $Q,$ then the locus of the mid-point of $PQ$ is

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