Question

If the line $y=7x-25$ meets the circle $x^2+y^2=25$ in points $A$, $B$, then the distance between $A$ and $B$ is

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

Answer 1

The correct option is C

$5\sqrt{2}$

Explanation for the correct option:

Step 1: Find the point of intersection.

In the question, an equation of the circle $x^2+y^2=25$ and an equation of the line $y=7x-25$ is given.

Find the point of intersection of the given circle and line as follows:

$$\begin{gathered} x^2+{(7x-25)}^2=25 \\ \Rightarrow x^2+49x^2+625-350x=25 \\ \Rightarrow 50x^2-350x+600=0 \\ \Rightarrow x^2-7x+12=0 \\ \Rightarrow x=\frac{7\pm \sqrt{{\left(-7\right)}^2-4\left(12\right)}}{2}\left\{x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\right\} \\ \Rightarrow x=\frac{7\pm \sqrt{49-48}}{2} \\ \Rightarrow x=\frac{7\pm 1}{2} \\ \Rightarrow x=3,4 \end{gathered}$$

Find the value of $y$ when $x=3$ by substituting $x=3$ in the given equation of the line.

$$\begin{gathered} y=7\left(3\right)-25 \\ \Rightarrow y=-4 \end{gathered}$$

Find the value of $y$ when $x=4$ by substituting $x=4$ in the given equation of the line.

$$\begin{gathered} y=7\left(4\right)-25 \\ \Rightarrow y=3 \end{gathered}$$

Therefore, the point of intersections $A$ and $B$ are $(3,-4)$ and $(4,3)$.

Step 2: Find the distance between the points of intersection.

We know that if the two points are $(x_1,y_1)$ and $(x_2,y_2)$. Then the distance $d$ between the points is given by: $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$.

So, the distance $d$ between the points of intersection is:

$$\begin{gathered} d=\sqrt{{\left(4-3\right)}^2+(3-{(-4))}^2} \\ \Rightarrow d=\sqrt{1^2+7^2} \\ \Rightarrow d=\sqrt{50} \\ \Rightarrow d=5\sqrt{2} \end{gathered}$$

Therefore, the distance between the points of intersection $A$ and $B$ is: $5\sqrt{2}$.

Hence, option (C) is the correct answer.