$C_1$ is the circle with centre at $O(0,0)$ and radius $4, C_2$ is a variable circle with centre at $(\alpha…

$C_1$ is the circle with centre at $O(0,0)$ and radius $4, C_2$ is a variable circle with centre at $(\alpha, \beta)$ and radius 5 . If the common chord of $\mathrm{C}_1$ and $\mathrm{C}_2$ has slope $\frac{3}{4}$ and of maximum length, then one of the possible values of $\alpha+\beta$ is
  1. $\frac{21}{5}$
  2. $\frac{3}{5}$
  3. $\frac{1}{5}$
  4. $\frac{19}{5}$

Solution

Equation of circle $C_1$ be $x^2+y^2=16$ Equation of circle $C_2$ be $(x-\alpha)^2+(y-\beta)^2=25$ $\Rightarrow x^2-2 \alpha x+\alpha^2+y^2-2 \beta y+\beta^2=25$
Equation of common chord is $C_2-C_1=0$ $\Rightarrow-2 \alpha x-2 \beta y+\alpha^2+\beta^2=9$
Slope of common chord $=\frac{-\alpha}{\beta}=\frac{3}{4}$ Let $\alpha=-3 \lambda$ and $\beta=4 \lambda$ Maximum length of common chord is diameter of smaller circle $C_1$. So, chord passes through centre of $C_2(0,0)$. $\begin{aligned} & \therefore \alpha^2+\beta^2=9 \Rightarrow 9 \lambda^2+16 \lambda^2=9 \Rightarrow \lambda=\frac{3}{5} \\ & \Rightarrow \alpha+\beta=\frac{-9}{5}+\frac{12}{5}=\frac{3}{5} \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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