Question

If the line $2y=4x-6$ cuts the curve $y^2=16x$ at points $A$ and $B,$ then the equation of circle having $AB$ as diameter is

• A. $x^2+y^2-x+8y+\frac{87}{4}=0$
• B. $x^2+y^2-7x-8y-\frac{87}{4}=0$
• C. $x^2+y^2+x+8y+\frac{105}{4}=0$
• D. $x^2+y^2+x+8y-\frac{105}{4}=0$

Answer 1

The correct option is B $x^2+y^2-7x-8y-\frac{87}{4}=0$

Given equations are $2y=4x-6$ and $y^2=16x$
As the line and curve are intersecting at points $A$ and $B,$
solving them, we get
$(2x-3{)}^2=16x\Rightarrow x^2-7x+\frac{9}{4}=0\cdots (1)$
This is quadratic equation in $x$.

Again, solving both the equations, we get
$y^2=16\times \left(\frac{y+3}{2}\right)\Rightarrow y^2-8y-24=0\cdots \left(2\right)$
This is quadratic equation in $y$.

Therefore, the equation of circle whose diameter is $AB$ is given by adding equations $\left(1\right)$ and $\left(2\right)$
$x^2+y^2-7x-8y-\frac{87}{4}=0$