Question

Locus of the mid-point of chord of circle \(x^2 + y^2 - 6x - 8y - 11 = 0\), subtending a right angle at the center is

• A. \(x^2 + y^2 - 6x - 8y + 7 = 0\)
• B. \(x^2 + y^2 - 6x - 8y - 7 = 0\)
• C. \(x^2 + y^2 + 6x + 8y - 7 = 0\)
• D. \(x^2 + y^2 - 6x + 8y + 7 = 0\)

Answer 1

The Correct Option is B

Step 1: Identifying the Circle's Center and Radius:
Rewrite $x^2+y^2-6x-8y-11=0$ in standard form by completing the square: \[ (x-3)^2 + (y-4)^2 = 9 + 16 + 11 = 36 \] So the center is $C = (3,4)$ and the radius is $r = 6$.
\includegraphics[width=0.5\linewidth]{m10.png}
Step 2: Setting up the Geometry:
Let $M = (h,k)$ be the midpoint of a chord $AB$ that subtends a right angle at center $C$.
Since $AB$ subtends a right angle at $C$, we have $\angle ACB = 90°$.
In the right triangle $ACB$, $|CA| = |CB| = r = 6$ (radii) and $\angle ACB = 90°$.
The midpoint $M$ of the hypotenuse $AB$ satisfies: \[ |CM| = \frac{|AB|}{2} \] Also, since $\triangle ACB$ is isoceles right-angled at $C$: \[ |AB|^2 = |CA|^2 + |CB|^2 = 36 + 36 = 72 \] But more directly: the perpendicular from the center to a chord bisects it, and $\angle ACM = 45°$ (half of $90°$), so: \[ |CM| = |CA|\cos 45° = 6 \cdot \frac{1}{\sqrt{2}} = 3\sqrt{2} \] Step 3: Writing the Locus Equation:
\[ |CM|^2 = (3\sqrt{2})^2 = 18 \] \[ (h-3)^2 + (k-4)^2 = 18 \] \[ h^2 - 6h + 9 + k^2 - 8k + 16 = 18 \] \[ h^2 + k^2 - 6h - 8k + 25 = 18 \] \[ h^2 + k^2 - 6h - 8k + 7 = 0 \] Step 4: Final Answer:
The locus is $x^2 + y^2 - 6x - 8y + 7 = 0$.
The answer is Option (1).