A straight line cuts the circle $x^{2} + y^{2} = 25$ in two distinct points $A$ and $B$ . The distance of $A$ and $B$ is 2 from the point $(3, 4)$ . Find the equation of the line $A B$ .

3 x + 4 y = 23

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4 x - 3 y = 0

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3 x + 4 y - 15 = 0

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4 x + 3 y = 14

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Solution

The correct option is A $3 x + 4 y = 23$
The locus of the points at a distance of $2$ from $(3, 4)$ would be a circle with radius 2 and centre as $(3, 4)$
The equation of the circle is -
${(x - 3)}^{2} + {(y - 4)}^{2} = 4$
This circle will intersect the circle $x^{2} + y^{2} = 25$ at two points. We can find the equation of the common chord by subtracting the two equations of the circles.
$∴$ The equation of common chord is :
$(x^{2} - (x^{2} - 6 x + 9)) + (y^{2} - (y^{2} - 8 y + 16)) = 25 - 4$
On simplifying we get,
$6 x + 8 y = 46$
$\Rightarrow 3 x + 4 y = 23$
Hence, option A is correct

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A straight line cuts the circle x2+y2=25 in two distinct points A and B. The distance of A and B is 2 from the point (3,4). Find the equation of the line AB.

A straight line cuts the circle $x^{2} + y^{2} = 25$ in two distinct points $A$ and $B$ . The distance of $A$ and $B$ is 2 from the point $(3, 4)$ . Find the equation of the line $A B$ .