Let the circles C 1 : x 2 + y 2 = 9 and C 2 : x - 3 2 + y - 4 2 = 16 , intersect at the points X and Y .…

1. $\left(II\right)-\left(T\right)$
2. $\left(I\right)-\left(S\right)$
3. $\left(I\right)-\left(U\right)$
4. $\left(II\right)-\left(Q\right)$

Solution

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) $\alpha =\frac{10}{3}$

Asked in: JEE Advanced 2019 (Paper 2)

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