Question

Angle between tangents drawn to $x^2+y^2-2x-4y+1=0$ at the points where it is cut by the line $y=2x+c$ is $\frac{\pi }{2}$, then

• A. $|c|=\sqrt{5}$
• B. $|c|=2\sqrt{5}$
• C. $|c|=\sqrt{10}$
• D. $|c|=2\sqrt{10}$

Answer 1

The correct option is D $|c|=2\sqrt{10}$

$ABisy=2x+cEquationofchordofcontactABfromp(h,k)is:\Rightarrow T=S,\Rightarrow x\left(h\right)+y\left(k\right)-(x+h)-2(y+k)+1=h^2+k^2-2h-4k+1\Rightarrow x(h-1)+y(k-2)+(-h^2+h+2k-k^2)=0\Rightarrow y=\frac{x(1-h)}{k-2}+\frac{(h^2+k^2-h-2k)}{k-2}Comparingwithy+2x+c.\Rightarrow \frac{1-h}{k-2}=2\Rightarrow 1-h=2k-4.\Rightarrow 2k+h=5\to \left(1\right).and\frac{h^2+k^2-(h+2k)}{k-2}=c\Rightarrow \frac{h^2+k-5}{k-2}=c\to \left(2\right).In△PAM\&PMB:\Rightarrow PA=PB\left(Tangentfromsameexternalpoint\right)\Rightarrow PM=PM\left(Common\right)\angle PMA=\angle PMB={90}^0\therefore △PAM\cong △PMB\therefore \angle APM=\angle MPB={45}^{0}In△OAP:\Rightarrow sin45=\frac{OA}{OP}\Rightarrow \frac{1}{\sqrt{2}}=\frac{r}{OP}r=\sqrt{1+4-1}=2.\Rightarrow \frac{1}{\sqrt{2}}=\frac{2}{\sqrt{(h-1{)}^2+(k-2{)}^2}}\Rightarrow (h-1{)}^2+(k-2{)}^2=8andh=5-2k\to from\left(1\right)\Rightarrow (4-2k{)}^2+(k-2{)}^2=8\Rightarrow 4(k-2{)}^2+(k-2{)}^2=8\Rightarrow 5(k-2{)}^2=8\Rightarrow k-2\frac{\pm 2\sqrt{2}}{\sqrt{5}}.\Rightarrow k-2=\pm 2\sqrt{\frac{2}{5}}.\Rightarrow k=2\pm 2\sqrt{\frac{2}{5}}.h=5-2k=5-(4\pm 4\sqrt{\frac{2}{5}}).=1\mp 4\sqrt{\frac{2}{5}}.Puttingin(2):\Rightarrow C=[1+16\left(\frac{2}{5}\right)\pm 8\sqrt{\frac{2}{5}}+(4+4(\frac{2}{5})\pm 8\sqrt{\frac{2}{5}})-5]\pm 2\sqrt{\frac{2}{5}}C=\frac{1+32\mp 8\sqrt{\frac{2}{5}}+4+\frac{8}{5}\pm 8\sqrt{\frac{2}{5}}-5}{\pm 2\sqrt{\frac{2}{5}}}=\frac{-4+4+8}{2\sqrt{\frac{2}{5}}}=4\sqrt{\frac{5}{2}}=\frac{4\sqrt{10}}{2}=2\sqrt{10}.Ans.$