The distance of a point $P(h,k)$

Question

The distance of a point $P(h,k)$ from a pair of lines passing through the origin is $d$ units. Prove that the equation of the pair of lines is $(xk-hy)^2=d^2(x^2+y^2)$.

Please help me. I am having trouble in starting.

Help much appreciated

Answer 1

Consider the circle with center $(h,k)$ and radius $d$. The equation is \begin{align*} (x-h)^2 + (y-k)^2 - d^2 &= 0\\ x^2+y^2-2hx - 2ky + h^2 +k^2 - d^2 &= 0 \end{align*} We want the pair of tangents to this circle from the origin. The general form of the pair of tangents from $(x_1,y_1)$ to a second degree curve $S=0$ is $T^2 = SS_1$ where $T$ is the tangent expression for the curve. In this case, \begin{align*} T &= xx_1 + yy_1 -h(x+x_1) - k(y+y_1) + h^2 + k^2-d^2\\ S &= x^2+y^2-2hx - 2ky + h^2 +k^2 - d^2\\ S_1&= h^2+k^2-d^2 \end{align*} where $(x_1,y_1) = (0,0)$. Thus the required equation is \begin{align*} (hx +ky -(h^2+k^2-d^2))^2 = (x^2+y^2-2hx - 2ky + h^2 +k^2 - d^2)(h^2+k^2-d^2) \end{align*} and this simplifies to \begin{align*} (kx-hy)^2 &= d^2(x^2+y^2) \end{align*}

Another way to solve the problem: The circle with center $(h,k)$ and radius $d$ is \begin{align*} x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - d^2 = 0 \end{align*}
The circle with $(0,0)$ and $(h,k)$ as ends of diameter is \begin{align*} x(x-h)+y(y-k) = 0 \end{align*} The common chord of these circles is \begin{align*} hx + ky = h^2+k^2-d^2 \end{align*} If this common chord meets the circles at $A,B$, then the lines $OA, OB$ are tangents to the first circle and hence they are at a distance $d$ from the point $(h,k)$. We want to find the pair of lines joining the origin to the points of $$ x^2 + y^2 - 2(hx+ky)\left(\frac{hx+ky}{h^2 + k^2 - d^2}\right) + (h^2+k^2-d^2)\left(\frac{hx+ky}{h^2 + k^2 - d^2}\right)^2 = 0 $$ This simplifies to $$ (kx-hy)^2 = d^2(x^2+y^2) $$

Answer 2

Let equation of any of the lines be $y-mx=0$. From the given condition (using distance formula) $\dfrac{|k-mh|}{\sqrt{1+m^2}}=d$. Since this is quadratic in $m$ its roots will give the two slopes of required lines. All we have to do is to eliminate $m$ from this. $m=y/x$. Thus it becomes$(k-mh)^2=d^2(1+m^2)$, substituting $m$ we get $(k-\frac yx h)^2=d^2(1+{(\frac yx )}^2)\Rightarrow (kx-hy)^2=d^2(x^2+y^2)$.