Question

The equation of the circle concentric with $x^2-3x+4y-c=0$ and passing through (-1, -2) is

• A. $x^2+y^2-3x+4y-1=0$
• B. $x^2+y^2-3x+4y=0$
• C. $x^2+y^2-3x+4y+2=0$
• D. none of these

Answer 1

The correct option is B

$x^2+y^2-3x+4y=0$

The centre of the circle $x^2+y^2-3x+4y-c=0$
is $(\frac{3}{2},-1)$

Therefore, the centre of the required circle is
$(\frac{3}{2},-2)$

${(x-\frac{3}{2})}^2+(y+2{)}^2=a^2\dots (1)$

Also, circle (i) passes through (-1, -2)

$\therefore {(-1-1\frac{3}{2})}^2+(-2+2{)}^2=a^2$

$\Rightarrow a=\frac{5}{2}$

Substituting the value of a in equation (1):

${(x-\frac{3}{2})}^2+(y+2{)}^2={\left(\frac{5}{2}\right)}^2$

$\Rightarrow \frac{(2x-3{)}^2}{4}+(y+2{)}^2=25$

$\Rightarrow x^2+y^2-3x+4y=0$

Hence the required equation of th circle =0