Question

Let a line perpendicular to the line $2x - y = 10$ touch the parabola $y^2 = 4(x - 9)$ at the point $P$. The distance of the point $P$ from the centre of the circle $x^2 + y^2 - 14x - 8y + 56 = 0$ is _____.

Answer 1

Given:

$y^2 = 4(x-9)$,

which represents a parabola with vertex at $(9,0)$ and axis along the $x$-axis.

Step 1: Finding the Slope of the Perpendicular Line The given line is:

$2x - y = 10.$

Rearranging:

$y = 2x - 10,$

with a slope of $2$. A line perpendicular to this has a slope:

$m = -\frac{1}{2}.$

Step 2: Equation of the Tangent The equation of the tangent to the parabola $y^2 = 4(x-9)$ at a point $(x_1, y_1)$ is given by:

$yy_1 = 2(x + x_1 - 9).$

Substituting the slope $m = -\frac{1}{2}$ into the equation of the tangent:

$y = -\frac{1}{2}x + c.$

Equating with the general form and solving for the point of contact $P$, we find:

$P(13, -4).$

Step 3: Centre of the Circle Given the equation of the circle:

$x^2 + y^2 - 14x - 8y + 56 = 0.$

Completing the square:

$(x-7)^2 + (y-4)^2 = 9.$

The centre of the circle is:

$C(7, 4).$

Step 4: Calculating the Distance $CP$ The distance between point $P(13, -4)$ and the centre $C(7, 4)$ is given by:

$CP = \sqrt{(13-7)^2 + (-4-4)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10.$

Therefore, the distance $CP$ is $10$.