Question

The line 2x + y + 1 = 0 intersects the circle $x^2 + y^2 = 9$ at two distinct points A and B and the line 3x + ky - 4 = 0 intersects the circle $x^2 + y^2 - 2x + 4y + 1 = 0$ at C and D, then

• A. if A, B, C, D are concyclic, then k is -1.
• B. if A, B, C, D are concyclic, then k is 1.
• C. if a circle S passes through A, B, C, D, then radical centre of three circles is $\left(\frac{3}{5}, -\frac{11}{5}\right)$
• D. if a circle S passes through A, B, C, D, then radical centre of three circles is (-2, 1).

Answer 1

Answer: (A), (C)

Options (A) and (C) are correct.

Answer 2

Solution:

We have two given circles and two lines:

1. Circle 1:

   $$x^2+y^2=9,$$

   intersected by line

   $$L_1: 2x+y+1=0.$$

   So the points $A$ and $B$ are the intersections of Circle 1 and $L_1$. Thus, if a circle $S$ passes through $A$ and $B$, the common chord of Circle 1 and $S$ is $L_1$. Therefore, the radical axis of Circle 1 and $S$ is

   $$2x+y+1=0.$$

2. Circle 2:

   $$x^2+y^2-2x+4y+1=0,$$

   intersected by line

   $$L_2: 3x+ky-4=0.$$

   So the points $C$ and $D$ are given by the intersections of Circle 2 and $L_2$. Hence if $S$ passes through $C, D$ also, the radical axis of Circle 2 and $S$ is

   $$3x+ky-4=0.$$

Also, the radical axis of Circle 1 and Circle 2 can be found by subtracting their equations. Write Circle 1 as

$$x^2+y^2-9=0,$$

and Circle 2 as

$$x^2+y^2-2x+4y+1=0.$$

Subtracting:

$$\bigl[x^2+y^2-9\bigr] - \bigl[x^2+y^2-2x+4y+1\bigr] =0,$$ $$2x-4y-10=0 \quad \Longrightarrow \quad x-2y-5=0.$$

Thus the radical axis of Circle 1 and Circle 2 is

$$x-2y-5=0.$$

Now, if the four points $A, B, C, D$ are concyclic, then there is a circle $S$ through them. The three circles (Circle 1, Circle 2 and $S$) have pairwise radical axes which are:

• $L_1: 2x+y+1=0$ (from Circle 1 and $S$),
• $L_2: 3x+ky-4=0$ (from Circle 2 and $S$),
• $L_3: x-2y-5=0$ (from Circle 1 and Circle 2).

By the Radical Axis Theorem these three lines must be concurrent.

Step 1. Find the intersection of $L_1$ and $L_3$:

$$L_1:\; 2x+y+1=0 \quad\Longrightarrow\quad y=-2x-1.$$

Substitute into $L_3:$

$$x-2(-2x-1)-5= x+4x+2-5=5x-3=0,$$ $$\Rightarrow x=\frac{3}{5},\quad y=-2\left(\frac{3}{5}\right)-1=-\frac{6}{5}-1=-\frac{11}{5}.$$

Thus, $L_1$ and $L_3$ meet at $\left(\frac{3}{5},-\frac{11}{5}\right)$.

Step 2. This intersection must also lie on $L_2:$

Substitute $x=\frac{3}{5}$ and $y=-\frac{11}{5}$ into $L_2:$

$$3\cdot\frac{3}{5}+k\left(-\frac{11}{5}\right)-4= \frac{9}{5}-\frac{11k}{5}-4=0.$$

Multiply by 5:

$$9-11k-20=0 \quad \Longrightarrow \quad -11k-11=0,$$ $$\Rightarrow k=-1.$$

Thus, when the four points are concyclic, we must have $k=-1$.

Radical centre of the three circles:

The radical centre is the common point of the three radical axes. We already computed the intersection of $L_1$ and $L_3$ to be $\left(\frac{3}{5}, -\frac{11}{5}\right)$. This is the desired radical centre.

Therefore:

• (A) is true since concyclicity forces $k=-1$.
• (B) is false.
• (C) is true (the radical centre is $\left(\frac{3}{5}, -\frac{11}{5}\right)$).
• (D) is false.