Question

The equation of circle touching the line $2x+3y+1=0$ at the point $(1,-1)$ and passing through the focus of the parabola $y^2=4x$ is

• A. $3x^2+3y^2-8x+3y+5=0$
• B. $3x^2+3y^2+8x-3y+5=0$
• C. $x^2+y^2-3x+y+6=0$
• D. none of these

Answer 1

Answer: (A)

$3x^2+3y^2-8x+3y+5=0$

Let equation of circle

$(x-h{)}^2+(y-k{)}^2=r^2\left(1\right)$

Since, $2x+3y+1=0$ is tangent to circle with slope $\left(\frac{-2}{3}\right)$ and intersecting point $(1,-1)$

The line passing through $(1,-1)$ and perpendicular to $2x+3y+1=0$ must pass through center.

Slope of perpendicular line $m_1{m}_2=-1m_1=\frac{3}{2}$

Equation of center line

$(y+1)=\frac{3}{2}(x-1)2y+5=3x$

Center is $(h,k)2k+5=3h\left(2\right)$

Circle also passes through $(1,-1)$ and focus of $(y^2=4x)$ i.e. $(1,0)$

Putting $(1,-1)$ in equation $\left(1\right)$

$(1-h{)}^2+(-1-k{)}^2=r^2\left(3\right)$

Putting $(1,0)$ in equation $\left(1\right)$

$(1-h{)}^2+(-k{)}^2=r^2\left(4\right)$

Subtracting $\left(3\right)$ from $\left(4\right)$

$(k+1{)}^2-k^2=0\Rightarrow k=\frac{-1}{2}$

Putting this value of $k$ in equaiton $\left(2\right)$

$2\left(\frac{-1}{2}\right)+5=3h\Rightarrow h=\frac{4}{3}$

Radius of circle is distance of center from line $2x+3y+1$

$r=\left|\frac{2\left(\frac{4}{3}\right)+3\left(\frac{-1}{2}\right)+1}{\sqrt{13}}\right|=\left|\frac{16-9+6}{6\sqrt{13}}\right|=\frac{\sqrt{13}}{6}$
Equation of circle

${(x-\frac{4}{3})}^2+{(y-\frac{1}{2})}^2=\frac{13}{36}3x^2+3y^2-8x+3y+5=0$