Question

Extremities of a diagonal of a rectangle are $(0,0)$ and $(4,3)$. The equation of the tangent to the circumcircle of the rectangle which are parallel to this diagonal are

• A. $16x+8y+25=0$
• B. $6x-8y+25=0$
• C. $8x+6y-25=0$
• D. $6x-8y-25=0$

Answer 1

Answer: (B), (D)

$6x-8y-25=0$
$6x-8y+25=0$

Extremities of the diagonal $OA$ of the rectangle are $O(0,0)$ and $A(4,3)$.

Then, $OA$ is the diameter of the circumcircle, so equation of the circumcircle is

$x(x-4)+y(y-3)=0\Rightarrow x^2+y^2-4x-3y=0$

$\Rightarrow {(x-2)}^2+{(y-\frac{3}{2})}^2={\left(\frac{5}{2}\right)}^2$ ...(1)

$m=$ slope of $OA=\frac{3}{4}$ ...(2)

$\therefore$ Tangent parallel to the diagonal $OA$ are

$y-\frac{3}{2}=\frac{3}{4}(x-2)\pm \frac{5}{2}\sqrt{1+\frac{9}{16}}\Rightarrow 6x-8y\pm 25=0$