Question

If the line $y=7x-25$ meets the circle$x^2+y^2=25$ in the 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 answer:

The equation of the given line is

$y=7x-25$$...\left(i\right)$

The equation of the given circle is

$x^2+y^2=25$$...\left(ii\right)$

By solving the equation of the line and the circle simultaneously we get the co-ordinates of the points of intersection

$\Rightarrow$ $x^2+{\left(7x-25\right)}^2=25$ …[From(i) and (ii)]

$\Rightarrow$$x^2+49x^2-350x+625=25$

$\Rightarrow$ $50x^2-350x+600=0$

$\Rightarrow$ $x^2-7x+12=0$

$\Rightarrow$ $\left(x-3\right)\left(x-4\right)=0$

$\Rightarrow$ $x=4$ and $x=3$

When $x=4$, $y=7\left(4\right)-25$ When $x=3$, $y=7\left(3\right)-25$

$y=3$ $y=-4$

Hence, $A\left(4,3\right)$ and $B\left(3,-4\right)$ are the points of intersection of the line and the circle

The distance between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given as

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Hence, the distance between the points $A\left(4,3\right)$ and $B\left(3,-4\right)$ is given as

$\Rightarrow$$AB=\sqrt{{\left(4-3\right)}^2+{\left(3-(-4)\right)}^2}$

$\Rightarrow$ $=\sqrt{1+49}$

$\Rightarrow$ $=\sqrt{25\times 2}$

$\Rightarrow$$AB=5\sqrt{2}$ units.

Hence, the distance $AB$ is $5\sqrt{2}$ units, so option (C) is the correct answer.