Question

If the straight lines joining the origin to the points of intersection of the straight line $69x+25y=3450$ and the circle ${(x-25)}^2+{(y-69)}^2=c^2$ are at right angle, then $c^2$ is equal to

• A. $5340$
• B. $5358$
• C. $5372$
• D. $5386$

Answer 1

Answer: (D)

$5386$

Making the equation of the circle homogeneous with the help of the equation of the line, we get
$\Rightarrow x^2+y^2-2(25x+69y)\frac{(69x+25y)}{2\times 69\times 25}+\left[\right(69{)}^2+(25{)}^2-c^2]\left[\frac{69x+25y}{2\times 69\times 25}\right]=0$

$\Rightarrow {(2\times 69\times 25)}^2(x^2+y^2)-4\times 69\times \left(25{)}^{2}x\right(69x+25y)-4\times$

$(69{)}^2\times 25y(69x+25y)+\left[\right(69{)}^2+(25{)}^2-c^2]\left[\right(69{)}^2{x}^2+(25{)}^2{y}^2+2\times 69\times 25xy]=0$

which is the equation of the required lines.

Since they are at right angles,
Co-efficient of $x^2+$ Co-efficient of $y^2=0$

$\Rightarrow \left[\right(69{)}^2+(25{)}^2-c^2]\left[\right(69{)}^2+\left(25{)}^2\right]=0$

$\Rightarrow c^2=(69{)}^2+(25{)}^2=4761+625=5386$