Question

The locus of centre of a circle which passes through the origin and cuts off a length of 4 unit from the line $x=3$ is

• A. $y^2+6x=0$
• B. $y^2+6x=13$
• C. $y^2+6x=10$
• D. $x^2+6y=13$

Hint:

Use the geometric relation $R^2 = p^2 + (L/2)^2$ where $p$ is perpendicular distance to a line and $L$ is chord length.

Answer 1

The Correct Option is B

Step 1: General Circle
Let the center be $C(-g, -f)$. Circle passes through origin, so $R^2 = g^2 + f^2$.
Step 2: Chord from Line
Distance from center $(-g, -f)$ to line $x - 3 = 0$: $d = |-g - 3| = |g + 3|$.
Step 3: Relationship
In $\Delta ABC$ (see diagram), $R^2 = d^2 + (\text{half-chord})^2$.
Given chord $= 4$, half-chord $= 2$.
$g^2 + f^2 = (g + 3)^2 + 2^2$.
Step 4: Locus
$g^2 + f^2 = g^2 + 6g + 9 + 4 \Rightarrow f^2 = 6g + 13$.
Replacing center $(x, y)$ with $(-g, -f)$, the center is $(x_0, y_0) = (-g, -f)$. Thus $g = -x, f = -y$.
$(-y)^2 = 6(-x) + 13 \Rightarrow y^2 + 6x = 13$.
Final Answer: (b)