Question

Let the line $2x+3y=18$ meet the circle $x^2+y^2=25$ at P $(\alpha ,\beta )$ where $\alpha ,\beta \in N$. Let Q $(\beta ,-\alpha )$ be another point on the above circle. Then

• A. $PQ=5\sqrt{2}$
• B. $PQ=2\sqrt{5}$
• C. Tangent at P is perpendicular to the tangent at Q.
• D. Tangent at P is parallel to the tangent at Q.

Answer 1

Answer: (A), (C)

The correct options are
A $PQ=5\sqrt{2}$
C Tangent at P is perpendicular to the tangent at Q.

$2x+3y=18\Rightarrow y=\frac{18-2x}{3}$
$\therefore {\left(\frac{18-2x}{3}\right)}^2=25\Rightarrow x^2=3,\frac{33}{13}$
But $\alpha ,\beta \in N$, $\therefore x=3\Rightarrow y=4$
$\therefore P=(3,4)$, $Q=(4,-3)$
$\left(a\right)PQ=\sqrt{50}=5\sqrt{2}$
$\left(b\right)x^2+y^2=25\Rightarrow \frac{dy}{dx}=-\frac{x}{y}$
Slope of tangent at $P,m_1=-\frac{3}{4}$
at $Q,m_2=\frac{4}{3}$
$m_1{m}_2=-1\Rightarrow$ perpendicular