Question

Locus of image of (4, 6) about the family of lines $a(3x-4y+5)+b(x+y-3)=0$ is curve C:

• A. P-3; Q-2; R-1; S-4
• B. P-3; Q-1; R-2; S-4
• C. P-4; Q-2; R-1; S-3
• D. P-4; Q-1; R-5; S-3

Answer 1

Answer: (D)

P-4; Q-1; R-5; S-3

Answer 2

We first note that the given family of lines

$$a(3x-4y+5)+b(x+y-3)=0$$

represents the pencil of lines through the common point of

$$3x-4y+5=0\quad\text{and}\quad x+y-3=0.$$

Solving these we get

$$x+y=3,\quad 3x-4(3-x)+5=0\quad\Rightarrow\quad 7x-7=0\quad\Rightarrow\quad x=1,\;y=2.$$

Thus, all lines in the family pass through $Q=(1,2)$.

Let $P=(4,6)$. The image $P'$ of $P$ under reflection in a line $L$ through $Q$ is given by

$$P' = Q + \Bigl[\,2\,(P-Q)\cdot d\,\Bigr]\,d-(P-Q),$$

where $d=(\cos\theta,\sin\theta)$ is the unit direction vector along $L$. Writing $u=P-Q=(3,4)$ (with $|u|=5$) we have

$$P'-Q=2(u\cdot d)d-u.$$

Now choose coordinates rotated so that the new $x'$–axis is along $u$. In these coordinates we may write $u=(5,0)$ and, letting $t=\theta-\phi$ (where $\phi$ is the angle of $u$) we get

$$P'-Q = \bigl(10\cos^2t-5,\;10\cos t\sin t\bigr).$$

Using the double‐angle formulas

$$10\cos^2t=5(1+\cos2t),\quad 10\cos t\sin t=5\sin2t,$$

we have

$$P'-Q = \bigl(5\cos2t,\,5\sin2t\bigr).$$

As $t$ varies, the tip of $P'-Q$ describes the circle

$$X^2+Y^2=25.$$

Returning to the original coordinates, we see that the locus $C$ of the reflections is a circle with center $Q=(1,2)$ and radius 5:

$$(x-1)^2+(y-2)^2=25.$$

Once $C$ is found, one may determine the other circles (or loci) as follows:

(P) Director circle of $C$:

For any circle $(x-h)^2+(y-k)^2=r^2$, the director circle (locus of points from which the tangents are perpendicular) is given by

$$(x-h)^2+(y-k)^2=2r^2.$$

Thus, for $C$ with $r=5$ and center $(1,2)$, its director circle is

$$(x-1)^2+(y-2)^2=50.$$

Expanding,

$$x^2+y^2-2x-4y+5-50=0\quad\Rightarrow\quad x^2+y^2-2x-4y-45=0.$$

This matches option (4) in List II.

(Q) Image of $C$ about the line $x+y=0$:

Reflection about $x+y=0$ sends any point $(x,y)$ to $(-y,-x)$. Thus the center $Q=(1,2)$ goes to $(-2,-1)$ while the radius remains unchanged. Hence the image is

$$(x+2)^2+(y+1)^2=25.$$

Expanding,

$$x^2+y^2+4x+2y+4+1-25=0\quad\Rightarrow\quad x^2+y^2+4x+2y-20=0.$$

This is option (1).

(R) A circle that cuts $C$ orthogonally:

Two circles $(x-h_1)^2+(y-k_1)^2=r_1^2$ and $(x-h_2)^2+(y-k_2)^2=r_2^2$ cut orthogonally if

$$(h_1-h_2)^2+(k_1-k_2)^2=r_1^2+r_2^2.$$

A good candidate is to take a circle with center different from $Q$. Among the given options, consider

$$x^2+y^2+8x+4y+4=0.$$

Complete the square:

$$(x+4)^2+(y+2)^2=16.$$

Its center is $(-4,-2)$ and radius $4$. The distance between the centers $(1,2)$ and $(-4,-2)$ is

$$\sqrt{(1+4)^2+(2+2)^2}=\sqrt{5^2+4^2}=\sqrt{25+16}=\sqrt{41}.$$

Now,

$$\text{LHS}=(\sqrt{41})^2=41,\quad\text{RHS}=5^2+4^2=25+16=41.$$

Thus they are orthogonal. This is option (5).

(S) Locus of points from which the tangents to $C$ subtend an angle $60^\circ$:

If tangents from a point $T$ to a circle of radius $r$ subtend an angle $2\alpha$, then the distance $d$ from $T$ to the center satisfies

$$\sin\alpha=\frac{r}{d}.$$

Here, $2\alpha=60^\circ$ so $\alpha=30^\circ$ and $\sin30^\circ=\frac{1}{2}$. Hence,

$$\frac{1}{2}=\frac{5}{d}\quad\Rightarrow\quad d=10.$$

Thus, the required locus is

$$(x-1)^2+(y-2)^2=100.$$

Expanding,

$$x^2+y^2-2x-4y+5-100=0\quad\Rightarrow\quad x^2+y^2-2x-4y-95=0.$$

This is option (3).

Now comparing with the given answer choices:

• (P) corresponds to option (4).
• (Q) corresponds to option (1).
• (R) corresponds to option (5).
• (S) corresponds to option (3).

Thus the correct matching is:

P-4; Q-1; R-5; S-3