Prove that the centres of the three circles
$x^{2} + y^{2} - 4 x - 6 y — - 12 = 0, x^{2} + y^{2} + 2 x + 4 y - 10 = 0$
and $x^{2} + y^{2} - 10 x - 16 y - 1 = 0$
are collinear.

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Solution

The given equation of circle are
$x^{2} + y^{2} - 4 x - 6 y - 12 = 0 \dots (i) x^{2} + y^{2} + 2 x + 4 y - 10 = 0 \dots (i i) x^{2} + y^{2} - 10 x - 16 y - 1 = 0 \dots (i i i)$
Let $C_{1} C_{2}$ and $C_{3}$ are the centres of (i) (ii) and (iii)
$∴ C_{1} = (- g, - f) = (2, 3) C_{2} = (- g, - f) = (- 1, - 2) C_{3} = (- g, - f) = (5, 8)$
$C_{1} C_{2} a n d C_{3}$ will be collinear if ar $(Δ C_{1} C_{2} C_{3}) = 0$
ar $(Δ C_{1} C_{2} C_{3}) = \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣$

$= \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 2 & 3 & 1 - 1 & - 2 & 1 5 & 8 & 1 \end{matrix} ∣ ∣ ∣ ∣$

$= \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 2 & 3 & 1 - 1 & - 2 & 1 5 & 8 & 1 \end{matrix} ∣ ∣ ∣ ∣$

$= \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 2 & 3 & 1 - 3 & - 5 & 0 3 & 5 & 0 \end{matrix} ∣ ∣ ∣ ∣ \begin{matrix} R_{2} \to R_{2} - R_{1} R_{3} \to R_{3} - R_{1} \end{matrix}$

$= \frac{1}{2} (- 15 + 15) = \frac{1}{2} \times 0 = 0$
$∴ C_{1} C_{2}$ and $C_{3}$ are collinear.

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x^{2} + y^{2} - 4 x - 6 y - 14 = 0

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