Question

If the line $y=4x-2$ cuts the curve $y^2=8x$ at points $A$ and $B,$ then the equation of circle having $AB$ as a diameter, is

• A. ${(x+\frac{1}{2})}^2+y^2-\frac{x}{2}-2y-4=0$
• B. ${(x-\frac{1}{2})}^2+y^2-\frac{x}{2}-2y-4=0$
• C. ${(x-\frac{1}{2})}^2+y^2+\frac{x}{2}-2y-4=0$
• D. ${(x-\frac{1}{2})}^2+y^2-\frac{x}{2}+2y-4=0$

Answer 1

The correct option is B ${(x-\frac{1}{2})}^2+y^2-\frac{x}{2}-2y-4=0$

For intersection of the given curve to the straight line, solving
$y=4x-2$ and $y^2=8x$
$\Rightarrow (4x-2{)}^2-8x=0$
$\Rightarrow 16{(x-\frac{1}{2})}^2-8x=0$
$\Rightarrow {(x-\frac{1}{2})}^2-\frac{x}{2}=0\cdots \left(1\right)$
This is a quadratic equation in $x$.

Now, $x=\frac{y+2}{4}$
Putting above value in $y^2=8x,$ we get
$y^2=8\times \frac{(y+2)}{4}$
$\Rightarrow y^2-2y-4=0\cdots \left(2\right)$
This is a quadratic equation in $y$.

Since, quadratic equation in $x+$quadratic equation in $y$ gives the equation of circle in diameter form,
adding equation $\left(1\right)$ and $\left(2\right),$ the required circle is
${(x-\frac{1}{2})}^2+y^2-\frac{x}{2}-2y-4=0$