Question

If the equation of a circle is $(4a - 3)^2 x^2 + a^2 y^2 + 6x - 2y + 2 = 0$, then its centre is:

• A. $(3, -1)$
• B. $(3, 1)$
• C. $(-3, 1)$
• D. $(-3, -1)$

Hint:

For a circle, $x^2$ and $y^2$ coefficients must be equal; then compare with $x^2 + y^2 + 2gx + 2fy + c = 0$ to find the centre.

Answer 1

The Correct Option is C

Step 1: For a general circle equation in standard form: \[ x^2 + y^2 + 2gx + 2fy + c = 0 \] The centre is given by: \[ (-g, -f) \] Step 2: In the given equation: \[ (4a - 3)^2 x^2 + a^2 y^2 + 6x - 2y + 2 = 0 \] To represent a circle, the coefficients of $x^2$ and $y^2$ must be equal, so: \[ (4a - 3)^2 = a^2 \] Step 3: Solve for $a$: Take square roots: \[ 4a - 3 = \pm a \] Case 1: $4a - 3 = a$ → $3a = 3$ → $a = 1$.
Case 2: $4a - 3 = -a$ → $5a = 3$ → $a = \frac35$. Step 4: For $a = 1$, equation becomes: \[ x^2 + y^2 + 6x - 2y + 2 = 0 \] Here, $2g = 6$ → $g = 3$, $2f = -2$ → $f = -1$.
Centre = $(-g, -f) = (-3, 1)$. Step 5: Therefore, the correct centre is $\mathbf(-3, 1)$.