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

Hint:In order to this question, to find the distance between $A\,and\,B$ , we will first substitute the value of $y$ in the equation of circle. Then we will find the value of both $x\,and\,y$ and then we can find the distance of $A\,and\,B$.

Complete step by step answer:
Given line is: $y = 7x - 25$ ….eq(i)
And the equation of the circle is ${x^2} + {y^2} = 25$ …..eq(ii)
Now we will substitute the equation(i) in the equation(ii):-
$\because {x^2} + {y^2} = 25 \\
\Rightarrow {x^2} + {(7x - 25)^2} = 25 \\
\Rightarrow {x^2} + 49{x^2} - 350x + 625 - 25 = 0 \\
\Rightarrow 50{x^2} - 350x + 600 = 0 \\
\Rightarrow {x^2} - 7x + 12 = 0 \\
\Rightarrow (x - 3)(x - 4) = 0 \\
\Rightarrow x = 3\,or\,x = 4 \\ $
Now, to find the value of $y$ , we will substitute the value of $x$ in equation(i):-
When $x = 3$ , $y = - 4$
When $x = 4$ , $y = 3$
So, the point where the given line meets the circle is $AB$ .
So, $A(3, - 4)\,and\,B(4,3)$ .
Now, we will find the distance of $AB$ :
$AB = \sqrt {{{(4 - 3)}^2} + {{(3 - ( - 4))}^2}} \\
\Rightarrow AB = \sqrt {1 + 49} \\
\Rightarrow AB = \sqrt {50} \\
\therefore AB = 5\sqrt 2 $
Therefore, the distance between $A\,and\,B$ is $5\sqrt 2 $ .

Hence, the correct option is C.

Note:A circle is a basic 2D shape which is measured in terms of its radius. The circles divide the plane into two regions such as interior and exterior regions. A circle is made up of all points in a plane that are evenly spaced from a fixed point. The fixed point is known as the circle's centre. The radius of the circle is the distance between the centre and any point on the circumference.