Question

If the length of the chord of the circle: $x^2+y^2=r^2\left(r>0\right)$ along the line $y-2x=3$ is $r$, then $r^2$ is equal to

• A. $12$
• B. $\frac{24}{5}$
• C. $\frac{9}{5}$
• D. $\frac{12}{5}$

Answer 1

The correct option is D

$\frac{12}{5}$

Step 1: Determine the points of intersections

The equation of the circle: $x^2+y^2=r^2\_\left(1\right)$.

The equation of the straight line: $y-2x=3$.

$\Rightarrow y=2x+3\_\left(2\right)$

Put $y=2x+3$ in equation $\left(1\right)$.

$$\begin{gathered} x^2+{\left(2x+3\right)}^2=r^2 \\ \Rightarrow x^2+4x^2+12x+9=r^2 \\ \Rightarrow 5x^2+12x+9-r^2=0 \end{gathered}$$

Solve the quadratic equation by the quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

$$\begin{gathered} \Rightarrow x=\frac{-12\pm \sqrt{{\left(12\right)}^2-4\left(5\right)\left(9-r^2\right)}}{2\left(5\right)} \\ \Rightarrow x=\frac{-12\pm \sqrt{144-20\left(9-r^2\right)}}{10} \\ \Rightarrow x=\frac{-12\pm 2\sqrt{36-5\left(9-r^2\right)}}{10} \\ \Rightarrow x=\frac{-6\pm \sqrt{5r^2-9}}{5} \end{gathered}$$

Put $x=\frac{-6\pm \sqrt{5r^2-9}}{5}$ in equation $\left(2\right)$.

$$\begin{gathered} \Rightarrow y=2\times \left(\frac{-6\pm \sqrt{5r^2-9}}{5}\right)+3 \\ \Rightarrow y=\frac{-12\pm 2\sqrt{5r^2-9}+15}{5} \\ \Rightarrow y=\frac{3\pm 2\sqrt{5r^2-9}}{5} \end{gathered}$$

Thus the points of intersects are $A\equiv \left(\frac{-6+\sqrt{5r^2-9}}{5},\frac{3+2\sqrt{5r^2-9}}{5}\right)$ and $B\equiv \left(\frac{-6-\sqrt{5r^2-9}}{5},\frac{3-2\sqrt{5r^2-9}}{5}\right)$.

Step 2: Determine the length of the chord

The length of the chord, $AB=\sqrt{{\left(\frac{-6+\sqrt{5r^2-9}}{5}-\frac{-6-\sqrt{5r^2-9}}{5}\right)}^2+{\left(\frac{3+2\sqrt{5r^2-9}}{5}-\frac{3-2\sqrt{5r^2-9}}{5}\right)}^2}$

$$\begin{gathered} \Rightarrow \sqrt{{\left(\frac{2\sqrt{5r^2-9}}{5}\right)}^2+{\left(\frac{4\sqrt{5r^2-9}}{5}\right)}^2}=r \\ \Rightarrow \sqrt{\frac{4\left(5r^2-9\right)}{25}+\frac{16\left(5r^2-9\right)}{25}}=r \\ \Rightarrow \sqrt{\frac{20\left(5r^2-9\right)}{25}}=r \\ \Rightarrow \sqrt{\frac{4\left(5r^2-9\right)}{5}}=r \\ \Rightarrow {\left(\sqrt{\frac{4\left(5r^2-9\right)}{5}}\right)}^2={\left(r\right)}^2 \\ \Rightarrow \frac{20r^2-36}{5}=r^2 \\ \Rightarrow 20r^2-36=5r^2 \\ \Rightarrow 15r^2=36 \\ \Rightarrow r^2=\frac{36}{15} \\ \Rightarrow r^2=\frac{12}{5} \end{gathered}$$

Hence, (D) is the correct option.