Mathematics - Circle Question with Solution | TestHub

Given centre of$C_1, C_2$and$C_3$are collinear hence$\left|\begin{array}{ccc}0& 0& 1\\ 3& 4& 1\\ h& k& 1\end{array}\right|=0$$3k=4h$…(i)$\Rightarrow MN$is diameter of$C_3$$MN=M{C}_1+C_1{C}_2+C_{2}N$$\begin{array}{cc}2r=r_1+C_1{C}_2+r_2& \left(\because MN=2r\right)\end{array}$$2r=3+\sqrt{{\left(3-0\right)}^2+{\left(4-0\right)}^2}+4$$\boxed{r=6}$…(ii)$\Rightarrow$Given$C_3$touches$C_1$at$\begin{array}{cc}M& \end{array}$So$\begin{array}{cc}\left|C_1{C}_3\right|=\left|r-3\right|& \end{array}$$h^2+k^2=9$…(iii) From (i) and (iii)$h=\pm \frac{9}{5}$and$k=\pm \frac{12}{5}$So centre of$C_3$will be$\left(\frac{9}{5},\frac{12}{5}\right)$$\Rightarrow$Now equation of common chord of$C_1$and$C_2$will be$C_1-C_2=0$$6x+8y=18$Equation of line$XY$is$3x+4y=9$…(iv) Distance of line$XY$from origin$C_{1}P=\frac{9}{5}$$\Rightarrow$now in$∆C_{1}Py$${\left(C_{1}P\right)}^2+{\left(PY\right)}^2={\left(C_{1}Y\right)}^2$$\begin{array}{cc}\frac{81}{25}+{\left(PY\right)}^2=\left(9\right)& \left(\because C_{1}Y=r_1=3\right)\end{array}$${\left(PY\right)}^2=\frac{144}{25}\Rightarrow PY=\frac{12}{5}$Length of$XY=2PY=\frac{24}{5}$$\Rightarrow$Line$ZW$is line$XY$Equation of$ZW=3x+4y=9$Distance of$C_3$from$ZW=\left|\frac{3\left(\frac{9}{5}\right)+4\left(\frac{12}{5}\right)-9}{5}\right|$$ZW=\frac{6}{5}$now$ZW=2\sqrt{{\left(6\right)}^2-{\left(\frac{6}{5}\right)}^2}$$ZW=\frac{24\sqrt{6}}{5}$$\left(\mathrm{I}\right) 2h+k=2\times \frac{9}{5}+\frac{12}{5}=\frac{30}{5}=6$$\left(\mathrm{I}\mathrm{I}\right)\frac{\mathrm{L}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }ZW}{\mathrm{L}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }XY}=\sqrt{6}$$\left(\mathrm{I}\mathrm{I}\mathrm{I}\right)\frac{\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }∆MZN}{\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }∆ZMW}=\frac{\frac{1}{2}\times MN\times PZ}{\frac{1}{2}\times ZW\times MP}$$=\begin{array}{cc}\frac{\frac{1}{2}MN\left(\frac{1}{2}ZW\right)}{\frac{1}{2}\times ZW\left(MG+GP\right)}& \left(\therefore PZ=\frac{1}{2}ZW\right)\end{array}$$=\begin{array}{cc}\frac{\frac{1}{2}\times 12\times \frac{12\sqrt{6}}{5}}{\frac{1}{2}\times \frac{24\sqrt{8}}{5}\left(\frac{24}{5}\right)}& \left(\begin{array}{c}\therefore M{C}_1=3\\ \therefore C_{1}A=\frac{9}{5}\end{array}\right)\end{array}$$=\frac{5}{4}$$\left(\mathrm{I}\mathrm{V}\right)$Common tangent to$C_1$and$C_3$is common chord to$C_1$and$C_3$$C_1-C_3=0$$3x+4y+15=0$Now this line is tangent to parabola$x^2=8\alpha y$$x^2=8\alpha \left(\frac{-3x-15}{y}\right)$$4x^2+24\alpha x+120\alpha =0$Apply$D=0$(for tangent, it will have repeated roots)$\alpha =\frac{10}{3}$