Question:

The locus of the centre of the circle, which cuts the circle $x^{2}+y^{2}-20 x+4=0$ orthogonally and touches the line $x=2$, is

Updated On: Aug 15, 2022
  • $x^{2}=16 y$
  • $y^{2}=4 x$
  • $y^{2}=16 x$
  • $x^{2}=4 y$
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The Correct Option is C

Solution and Explanation

Let the equation of circle be $x^{2}+y^{2}+2\, g x+2\, f \,y+c=0$ where, centre $(-g,-f)$ The centre of given circle $x^{2}+y^{2}-20 x+4=0$ is $(10,0)$ Condition of two circles cut. $\therefore 2\left(g_{1} \,g_{2}+f_{1}\, f_{2}\right) =c_{1}+c_{2} $ $2(-g \times 10+0 \times(-f)=c+4 $ $2(-10 \,g) =c+4$ Also, circle touch the line $x=2$. $\therefore$ The perpendicular distance from centre to the circle is equal to radius of the circle. $\therefore \frac{|-g-2|}{\sqrt{1}}=\sqrt{g^{2}+f^{2}-c}$ $\Rightarrow (g+2)=\sqrt{g^{2}+f^{2}-c}$ $\Rightarrow g^{2}+4+4 \,g=g^{2}+f^{2}-c$ $\Rightarrow f^{2}-4\, g-c-4=0$ $\Rightarrow f^{2}-4 \,g+4+20\, g-4=0$ $\Rightarrow f^{2}+16\, g=0$ Hence, the locus of $(-g,-f)$ is $y^{2}-16 x=0$ $ \Rightarrow y^{2}=16 x$
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Concepts Used:

Straight lines

A straight line is a line having the shortest distance between two points. 

A straight line can be represented as an equation in various forms,  as show in the image below:

 

The following are the many forms of the equation of the line that are presented in straight line-

1. Slope – Point Form

Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.

y – y0 = m (x – x0)

2. Two – Point Form

Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2)  are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes

The slope of P2P = The slope of P1P2 , i.e.

\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)

Hence, the equation becomes:

y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)

3. Slope-Intercept Form

Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by

y – c =m( x - 0 )

As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if

y = m x +c