Question

The co-ordinates of two points P and Q are $(x_1,y_1)$ and $(x_2,y_2)$ and O is the origin. If circles be described on OP and OQ as diameters then length of their common chord is

• A. $\frac{|x_1{y}_2+x_2{y}_1|}{PQ}$
• B. $\frac{|x_1{y}_2-x_2{y}_1|}{PQ}$
• C. $\frac{|x_1{x}_2+y_1{y}_2|}{PQ}$
• D. $\frac{|x_1{x}_2-y_1{y}_2|}{PQ}$

Answer 1

The correct option is B $\frac{|x_1{y}_2-x_2{y}_1|}{PQ}$

Let $S_1$ be the circle with OP as diameter.
$\Rightarrow S_1:x(x-x_1)+y(y-y_1)=0\Rightarrow x^2+y^2=x{x}_1+y{y}_1$
And $S_2$ be the circle with OQ as diameter.
$\Rightarrow S_1:x(x-x_2)+y(y-y_2)=0\Rightarrow x^2+y^2=x{x}_2+y{y}_2$
Also $PQ=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
Now equation of common chord is, $x{x}_1+y{y}_1=x{x}_2+y{y}_2\Rightarrow y=\frac{x_1-x_2}{y_2-y_2}x$
Solving this with $S_1$
$x^2+{\left(\frac{x_1-x_2}{y_2-y_2}x\right)}^2=x{x}_1+\frac{x_1-x_2}{y_2-y_2}x{y}_1=x\frac{y_1{x}_2-x_1{y}_2}{y_1-y_2}$
$\Rightarrow x=0,\frac{(y_1{x}_2-x_1{y}_2)(y_1-y_2)}{(x_1-x_2{)}^2+(y_1-y_2{)}^2}=0,\frac{(y_1{x}_2-x_1{y}_2)(y_1-y_2)}{P{Q}^2}$
Similarly $y=0,\frac{(y_1{x}_2-x_1{y}_2)(x_1-x_2)}{P{Q}^2}$
Thus end points of chord are $(0,0)$ and $(\frac{(y_1{x}_2-x_1{y}_2)(y_1-y_2)}{P{Q}^2},\frac{(y_1{x}_2-x_1{y}_2)(x_1-x_2)}{P{Q}^2})$
Hence length of the chord is $=\left|\frac{(y_1{x}_2-x_1{y}_2)\sqrt{(y_1-y_2{)}^2+(x_1-x_2{)}^2}}{P{Q}^2}\right|=\frac{|y_1{x}_2-x_1{y}_2|}{PQ}$