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CAT 2023 Slot 1 — QA Question 21

BasicsEasy

Let C be the circle x2 + y2 + 4x - 6y - 3 = 0 and L be the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to 60 degree. Then, the point at which L touches the line x = 6 is?

Answer & solution

  • A

    (6, 6)

  • B

    (6, 8)

  • (6, 3)

  • D

    (6, 4)

Solution

Easy

Complete the square to find the circle's centre and radius. From an external point LL, the two tangents make 60∘60^\circ between them, so each makes 30∘30^\circ with the line LOLO to the centre. That fixes the distance OLOL via a 3030-6060-9090 triangle; impose x=6x=6 on LL and solve for its yy.

1

Find centre and radius by completing the square:

x2+y2+4x−6y−3=0(x+2)2+(y−3)2=3+4+9=16=42\begin{aligned} &x^2+y^2+4x-6y-3=0\\ &(x+2)^2+(y-3)^2=3+4+9=16=4^2 \end{aligned}

Centre O=(−2,3)O=(-2,3), radius r=4r=4.

2

Use the tangent geometry. Let PP be a point of tangency. Triangle OPLOPL is right-angled at PP (radius ⊥\perp tangent), and LOLO bisects the 60∘60^\circ angle, so ∠OLP=30∘\angle OLP=30^\circ:

sin⁡30∘=OPOL=rOL ⇒ OL=rsin⁡30∘=41/2=8\sin 30^\circ=\frac{OP}{OL}=\frac{r}{OL}\ \Rightarrow\ OL=\frac{r}{\sin30^\circ}=\frac{4}{1/2}=8
3

Impose x=6x=6. Point L=(6,y)L=(6,y) sits at distance OL=8OL=8 from O=(−2,3)O=(-2,3):

(6−(−2))2+(y−3)2=8264+(y−3)2=64 ⇒ (y−3)2=0 ⇒ y=3\begin{aligned} &(6-(-2))^2+(y-3)^2=8^2\\ &64+(y-3)^2=64\ \Rightarrow\ (y-3)^2=0\ \Rightarrow\ y=3 \end{aligned}
4

Interpret. LL is the circle of radius 88 about OO; the horizontal distance from OO to x=6x=6 is exactly 88, so x=6x=6 is tangent to LL at the point level with the centre — confirming a single touch point.

O(-2,3) L(6,3) OL = 8 x = 6

LL touches x=6x=6 at (6, 3)\mathbf{(6,\,3)}.

CAT 2023 Slot 1 QA Q21: Let C be the circle x 2 + y 2 + 4x - 6y - 3 = 0 and L be the locus of the point of intersection of a pair of t — Solution | TheCATExam