Question

If OA and OB are two equal chords of the circle $x^2+y^2-2x+4y=0$ perpendicular to each other and passing through the origin, the slopes of OA and OB are the roots of the equation.

• A. $3m^2+8m-3=0$
• B. $3m^2-8m-3=0$
• C. $8m^2+3m-8=0$
• D. $8m^2+3m+8=0$

Answer 1

The correct option is B $3m^2-8m-3=0$

$\begin{array}{c}{x}^2+y^2-2x+4y=0\\ centre\left(1,-2\right)\\ OC=\sqrt{1^2+{\left(-2\right)}^2}=\sqrt{5}\\ OA\perp OB\Rightarrow OA=OB\left[given\right]\\ In\Delta AOB\\ O{A}^2+O{B}^2=A{B}^2\\ O{A}^2+O{A}^2={\left(2r\right)}^2\\ 2O{A}^2=4r^2\\ 2O{A}^2=20\\ \sqrt{O{A}^2}=\sqrt{10}\\ OA=\sqrt{10}\\ equationofOB=y=mx\&equationofOA=y=\frac{-x}{m}\\ OM=MB=OB=\sqrt{\frac{10}{2}}\\ In\Delta BOM,\\ C{M}^2=B{C}^2-B{M}^2\\ C{M}^2=\frac{5}{1}-\frac{10}{4}\\ C{M}^2=\frac{10}{4}\Rightarrow \frac{5}{2}\\ Accordingto\perp dis\tan ceformula\Rightarrow CM=\left|\frac{m+2}{\sqrt{m^2+1}}\right|\\ Cm\frac{{\left(m+2\right)}^2}{m^2+1}\Rightarrow \frac{m^2+4m+4}{m^2+1}=\frac{5}{2}\\ 2m^2+8m+8=5m^2+5\Rightarrow 3m^2-8m-3=0\\ m=3,\frac{-1}{3}\\ putm=3\\ equationofOB\Rightarrow y=mx\Rightarrow y=3xforOA\Rightarrow y=\frac{-x}{3}\\ m=\frac{1}{3}\\ equationofOBy=mx\Rightarrow y=\frac{-x}{3}forOAy=\frac{-x}{3}\end{array}$