Question

The tangent and normal at the point $P(18,12)$ of the parabola $y^2=8x$ intersect the $x-$axis at the point $Q$ and $R$ respectively. The equation of the circle through $P$, $Q$ and $R$ is given by

• A. $x^2+y^2-4x-396=0$
• B. $x^2+y^2-4x-360=0$
• C. $x^2+y^2-4x-346=0$
• D. $x^2+y^2-4x-300=0$

Answer 1

The correct option is A $x^2+y^2-4x-396=0$

$P(18,12)\Rightarrow P(\frac{c}{m},\frac{2a}{m})$

$\frac{2a}{m}=12\Rightarrow m=\frac{2a}{12}=\frac{2\times 2}{12}=\frac{1}{3}$

$y^2=8x\Rightarrow a=2$

Equation of tangent:

$y{y}_1=2a(x+x_1)$

$y\left(12\right)=2\times 2(x+18)$

$12y=4x+72$

Equation of normal:

$\frac{y-y_1}{y_1}=-\frac{(x-x_1)}{2a}$

$\frac{y-12}{12}=-\frac{(x-16)}{2\times 2}$

$\frac{y-12}{3}=-x+18$

$y=-3x+66$

These 2 equations $12y=4x+72$ and $y=-3x+66$, intersect at the points:

$P(18,12),Q(-18,0),R(22,0)$

Therefore the equation of circle passing through these points is:

$x^2+y^2-4x-396=0$

Because, when we substitute these points in the above circle equation, we get $0=0$,

These points satisfy the equation of circle, $x^2+y^2-4x-396=0$