If the equation of a circle is ^2+(2 -3)y^2-4x+6y-1=0, then the c… - Vidyadip

MCQ

If the equation of a circle is $\lambda\text{x}^2+(2\lambda-3)\text{y}^2-4\text{x}+6\text{y}-1=0,$ then the coordinates of centre are:

A
$\Big(\frac{4}{3},\ -1\Big)$
✓
$\Big(\frac{2}{3},\ -1\Big)$
C
$\Big(\frac{-2}{3},\ 1\Big)$
D
$\Big(\frac{2}{3},\ 1\Big)$

✓

Answer

Correct option: B.

$\Big(\frac{2}{3},\ -1\Big)$

To find the centre:
Coefficient of $x^2 =$ Coefficient of $y^2$
$\therefore\lambda=2\lambda-3$
$\Rightarrow\lambda=3$
Therefore, the given equation can be rewritten as $3\text{x}^2+3\text{y}^2-4\text{x}+6\text{y}-1=0.$
$\therefore\text{x}^2+\text{y}^2-\frac{4}{3}\text{x}+2\text{y}-\frac{1}{3}=0$
Thus, the coordinates of the centre is $\Big(\frac{2}{3},\ -1\Big).$

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