Question

The centre and radius of the circle \((a+1)x^2+3y^2-6x+9y+a+4 = 0\) are respectively ...

• A. \((-1,\frac{3}{2}),\frac{\sqrt{5}}{2}\)
• B. \((-1,-\frac{3}{2}),\frac{\sqrt{5}}{2}\)
• C. \((1,-\frac{3}{2}),\frac{\sqrt{5}}{2}\)
• D. \((1,\frac{3}{2}),\frac{\sqrt{5}}{2}\)

Hint:

A circle needs equal x^2 and y^2 coefficients, so a + 1 = 3.

Answer 1

The Correct Option is C

Step 1: Equal coefficients:
For a circle, $a+1=3$ so $a=2$. After dividing by 3 the constant term becomes $\tfrac{a+4}{3}=2$.

Step 2: Complete the square:
$x^2-2x+y^2+3y=-2$ gives $(x-1)^2+\left(y+\tfrac32\right)^2=-2+1+\tfrac94=\tfrac54$.

Step 3: Read off:
Centre $\left(1,-\tfrac32\right)$, radius $\sqrt{5/4}=\tfrac{\sqrt5}2$. Option (C).

Final Answer:
Completing the square gives the centre (1, -3/2). \[ \boxed{C} \]