Mathematics › Circle › Equation of Circle & its intercept
If the parametric values of two points $A$ and $B$ on the circles $x^2+y^2-6 x+4 y-12=0$ are $30^{\circ}$…
If the parametric values of two points $A$ and $B$ on the circles $x^2+y^2-6 x+4 y-12=0$ are $30^{\circ}$ and $90^{\circ}$ respectively, then the equation of chord $\mathbf{A B}$ is
$x+\sqrt{3} y=0$ $x-\sqrt{3} y=0$ $x+\sqrt{3} y-3(1+\sqrt{3})=0$ $\sqrt{3} x+\sqrt{3} y+61=0$ Solution
Given, equation of circle
$
x^2+y^2-6 x+4 y-12=0
$
$
(x-3)^2+(y+2)^2=(5)^2
$
parametric form of circle is
$
\begin{aligned}
& x=3+5 \cos \theta, y=-2+5 \sin \theta \\
& \text { When } \theta=30^{\circ} \\
& \quad A\left(3+5 \cos 30^{\circ},-2+5 \sin 30^{\circ}\right) \\
& \quad A\left(3+\frac{5 \sqrt{3}}{2},-2+\frac{5}{2}\right)=\left(\frac{6+5 \sqrt{3}}{2}, \frac{1}{2}\right)
\end{aligned}
$
$
\begin{aligned}
& \text { When } \theta=90^{\circ} \\
& \qquad \begin{array}{l}
B\left(3+5 \cos 90^{\circ},-2+5 \sin 90^{\circ}\right) \\
B(3,3)
\end{array}
\end{aligned}
$
$\therefore$ Equation of chord $A B$ where
$
\begin{aligned}
& A=\left(\frac{6+5 \sqrt{3}}{2}, \frac{1}{2}\right) \text { and } B=(3,3) \\
& y-3=\frac{\frac{1}{2}-3}{\frac{6+5 \sqrt{3}}{2}-3}(x-3) \\
\Rightarrow & y-3=\frac{-1}{\sqrt{3}}(x-3) \Rightarrow \sqrt{3} y-3 \sqrt{3}=-x+3 \\
\Rightarrow & x+\sqrt{3} y-3 \sqrt{3}-3=0 \\
\Rightarrow & x+\sqrt{3} y-3(1+\sqrt{3})=0
\end{aligned}
$
Asked in: AP EAMCET 2021 (25 Aug Shift 1)
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