Question

The equation of the circle described on the chord $3x+y+5=0$ of the circle $x^2+y^2=16$ as diameter is :

• A. $x^2+y^2+3x+y-11=0$
• B. $x^2+y^2+3x+y+1=0$
• C. $x^2+y^2+3x+y-2=0$
• D. $x^2+y^2+3x+y-22=0$

Answer 1

The correct option is A $x^2+y^2+3x+y-11=0$

$3x+y+5=0$ is a chord of circle $x^2+y^2=16$
$y=-5-3x$ ....... $\left(i\right)$
Put the value of $y$ in equation of circle, we get
$x^2+{(-5-3x)}^2=16$
$x^2+{9x}^2+25+30x=16$
${10x}^2+30x+9=0$
$x=\frac{-30\pm \sqrt{900-4\left(10\right)\left(9\right)}}{20}$
$x=\frac{-30\pm \sqrt{900-360}}{20}$
$x=\frac{-30\pm \sqrt{540}}{20}$
$x=\frac{-30\pm 6\sqrt{15}}{20}$
$x=\frac{-15\pm 3\sqrt{15}}{10}$
$y=-5-3\left[\frac{-15+3\sqrt{15}}{10}\right],-5-3\left[\frac{-15-3\sqrt{15}}{10}\right]$
$y=\frac{-5-9\sqrt{15}}{10},\frac{-5+9\sqrt{15}}{10}$
$\therefore$ End points of chord are $[\frac{-15+3\sqrt{15}}{10},\frac{-5-9\sqrt{15}}{10}],[\frac{-15-3\sqrt{15}}{10},\frac{-5+9\sqrt{15}}{10}]$
This chord acts as a diameter for another circle.
$\therefore$ Center = mid point of above points
$=[\frac{-30}{2\times 10},\frac{-10}{2\times 10}]=[\frac{-3}{2},\frac{-1}{2}]$
Equation of circle is $:{(x+\frac{3}{2})}^2+{(y+\frac{1}{2})}^2=r^2$ ....... $\left(ii\right)$
Radius = distance between centre & $[\frac{-15+3\sqrt{15}}{10},\frac{-5-9\sqrt{15}}{10}]$
$r^2={[\frac{-15+3\sqrt{15}}{10}+\frac{3}{2}]}^2+{[\frac{-5-9\sqrt{15}}{10}+\frac{1}{2}]}^2$
$r^2={\left[\frac{-15+3\sqrt{15}+15}{10}\right]}^2+{\left[\frac{-5-9\sqrt{15}+5}{10}\right]}^2$
$r^2=\frac{9\times 15}{100}+\frac{81\times 15}{100}$
$r^2=\frac{15}{100}\left[90\right]=\frac{27}{2}$
Put value of $r^2$ in $\left(ii\right)$, we get
${\left(x+\frac{3}{2}\right)}^2+{\left(y+\frac{1}{2}\right)}^2=\frac{27}{2}$
${4x}^2+9+12x+{4y}^2+1+4y=54$
${4x}^2+{4y}^2+12x+4y=44$
$x^2+y^2+3x+y-11=0$