CAT 2023 Slot 1 · Quant · Question 17

First put the circle in centre-radius form by completing the squares.

Grouping, \left({x}^{2}+4x \right)+\left({y}^{2}-6y \right)=3.

Completing each square, {\left( x+2 \right)}^{2}-4+{\left( y-3 \right)}^{2}-9=3.

So {\left( x+2 \right)}^{2}+{\left( y-3 \right)}^{2}=16, a circle with centre \left( -2,3 \right) and radius 4.

Now take an external point and draw the two tangents from it, with {60}^{∘} between them.

The line from that point to the centre bisects the angle, so each half-angle is {30}^{∘}.

In the right triangle formed by the point, the centre and a point of tangency, \sin{30}^{∘}=\frac{\text{radius}}{\text{distance to centre}}.

So the distance to the centre is \frac{4}{\sin{30}^{∘}}=\frac{4}{0.5}=8, the same for every such point.

Hence L is a circle concentric with C, centred at \left( -2,3 \right) with radius 8.

The distance from \left( -2,3 \right) to the vertical line x=6 is 6-\left( -2 \right)=8, exactly the radius — so L touches that line.

The point of contact is the foot of the perpendicular from the centre, which is \left( 6,3 \right) — the third option.

Shortcut. The whole question is one ratio: a {60}^{∘} angle between tangents doubles the radius, since \sin{30}^{∘}=\frac{1}{2}. The contact point then shares the centre's y-coordinate, which alone picks out option 3.