JEE Main 2021 — Mathematics Coordinate Geometry

Given the circle has centre $(2,3)$(2,3) and it passes through the point $(0,0).$(0,0).

So, the radius of the circle is $\sqrt{{2}^{2}+{3}^{2}}=\sqrt{13}$22+32​=13​

The equation of the line which passes through the point (0,0)&(2,3) is $y-0=\frac{3-0}{2-0}(x-0)\Rightarrow y=\frac{3}{2}x$y−0=2−03−0​(x−0)⇒y=23​x

So, the slope of $OC=\frac{3}{2}$OC=23​

From the diagram, the line $PQ$PQ is perpendicular to the line $OC.$OC.

So, the slope of $PQ$PQ is $-\frac{2}{3}$−32​

So, $\mathrm{tan}\theta =-\frac{2}{3}$tanθ=−32​

Using symmetric from of line

The coordinate of $P=(2+\sqrt{13}\mathrm{cos}\theta ,3+\sqrt{13}\mathrm{sin}\theta )$P=(2+13​cosθ,3+13​sinθ)

$$=(2+\sqrt{13}\cdot (-\frac{3}{\sqrt{13}}),3+\sqrt{13}(\frac{2}{\sqrt{13}}))$$=(2+13​⋅(−13​3​),3+13​(13​2​))

$$=(-1,5)$$=(−1,5)

The coordinate of $Q=(2-\sqrt{13}\mathrm{cos}\theta ,3-\sqrt{13}\mathrm{sin}\theta )$Q=(2−13​cosθ,3−13​sinθ)

$$=(2-\sqrt{13}\cdot (-\frac{3}{\sqrt{13}}),3-\sqrt{13}(\frac{2}{\sqrt{13}}))$$=(2−13​⋅(−13​3​),3−13​(13​2​))

$$=(5,1)$$=(5,1)