Question

Find the equation of the circle on the straight line joining the points of intersection of $a{x}^2+2hxy+b{y}^2=0$ and $lx+my=1$ as diameter.

Answer 1

Let the equations of given curve and straight line are

$a{x}^2+2hxy+b{y}^2=0$ ......$\left(1\right)$ and $lx+my=1$ ....$\left(2\right)$

Let $A(x_1,y_1)$ and $B(x_2,y_2)$ be the points of intersection of $\left(1\right)$ and $\left(2\right)$

From $\left(2\right)$ we have $my=1-lx$

or $y=\frac{1-lx}{m}$ .......$\left(3\right)$

Put the value of $y$ in $\left(3\right)$ in $\left(1\right)$ we get

$a{x}^2+2hx\left(\frac{1-lx}{m}\right)+b{\left(\frac{1-lx}{m}\right)}^2=0$

$\Rightarrow a{m}^2{x}^2+2hmx-2hml{x}^2+b+b{l}^2{x}^2-2lbx=0$

$\Rightarrow (a{m}^2-2hml+b{l}^2)x^2-2(lb-hm)x+b=0$ .....$\left(4\right)$

Clearly $x_1$ and $x_2$ are the roots of this equation

$\therefore$ Sum of the roots$=x_1+x_2=\frac{2(lb-hm)}{(a{m}^2-2hml+b{l}^2)}$ and product of the roots$=x_1{x}_2=\frac{b}{(a{m}^2-2hml+b{l}^2)}$ ......$\left(5\right)$

Eliminating $x$ from $\left(1\right)$ and $\left(2\right)$ by substituting $x=\frac{1-my}{l}$ in $\left(1\right)$ we get

$(a{m}^2-2hml+b{l}^2)y^2-2(am-hl)y+a=0$ ......$\left(6\right)$

Clearly $y_1$ and $y_2$ are the roots of this equation

$\therefore$ Sum of the roots$=y_1+y_2=\frac{2(am-hl)}{(a{m}^2-2hml+b{l}^2)}$ and product of the roots$=y_1{y}_2=\frac{a}{(a{m}^2-2hml+b{l}^2)}$ ......$\left(7\right)$

$\therefore$ equation of the circle with $AB$ as diameter is

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

$\Rightarrow x^2-(x_1+x_2)x+x_1{x}_2+y^2-(y_1+y_2)x+y_1{y}_2=0$

$\Rightarrow (x^2+y^2)-\frac{2(lb-hm)x}{(a{m}^2-2hml+b{l}^2)}+\frac{b}{(a{m}^2-2hml+b{l}^2)}-\frac{2(am-hl)y}{(a{m}^2-2hml+b{l}^2)}+\frac{a}{(a{m}^2-2hml+b{l}^2)}=0$

$\Rightarrow (x^2+y^2)(a{m}^2-2hml+b{l}^2)-2(lb-hm)x+2(am-hl)y+(a+b)=0$ is the required equation.