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: Understanding the Concept:
The general second degree equation represents a circle only if the coefficients of \(x^2\) and \(y^2\) are equal and there is no \(xy\) term.

Step 2: Find a:
Coefficients: \(a+1=3\), so \(a=2\). The equation is
\[ 3x^2+3y^2-6x+9y+6=0 \]

Step 3: Divide by 3:
\[ x^2+y^2-2x+3y+2=0 \]

Step 4: Centre and radius:
Compare with \(x^2+y^2+2gx+2fy+c=0\): \(g=-1\), \(f=\tfrac32\), \(c=2\). Centre \(=(-g,-f)=\left(1,-\tfrac32\right)\).
\[ r=\sqrt{g^2+f^2-c}=\sqrt{1+\tfrac94-2}=\sqrt{\tfrac54}=\tfrac{\sqrt5}{2} \]

Step 5: Choose:
Option (C). Options (A), (B) and (D) have a wrong sign for one coordinate of the centre.

Final Answer:
Centre (1, -3/2) and radius sqrt(5)/2. \[ \boxed{\left(1,-\tfrac32\right),\ \tfrac{\sqrt5}{2}} \]