Question

Paragraph for Question 46 to 47 Consider circles $C_1 : x^2 + y^2 - 4x - 6y - 3 = 0$ and $C_2 : x^2 + y^2 + 2x + 2y + 1 = 0$. Let $L_1 = 0$ represent the equation of one of the direct common tangent.

46. The equation of $L_1$ is

• A. x + 2 = 0
• B. x - 2 = 0
• C. 7x + 24y - 42 = 0
• D. 7x - 24y + 42 = 0

Answer 1

Answer: (A)

x + 2 = 0

Answer 2

The given circles are $C_1 : x^2 + y^2 - 4x - 6y - 3 = 0$ and $C_2 : x^2 + y^2 + 2x + 2y + 1 = 0$. The center and radius of $C_1$ are $O_1 = (2, 3)$ and $r_1 = \sqrt{2^2 + 3^2 - (-3)} = \sqrt{4+9+3} = \sqrt{16} = 4$. The center and radius of $C_2$ are $O_2 = (-1, -1)$ and $r_2 = \sqrt{(-1)^2 + (-1)^2 - 1} = \sqrt{1+1-1} = \sqrt{1} = 1$.

We need to find a direct common tangent $L_1$. Let's test the given options. (A) $L_1: x + 2 = 0$. Distance from $O_1(2,3)$ to $L_1$: $d_1 = \frac{|2+2|}{\sqrt{1^2+0^2}} = \frac{4}{1} = 4 = r_1$. Distance from $O_2(-1,-1)$ to $L_1$: $d_2 = \frac{|-1+2|}{\sqrt{1^2+0^2}} = \frac{1}{1} = 1 = r_2$. For $x+2=0$, the value of $x+2$ for $O_1$ is $2+2=4>0$ and for $O_2$ is $-1+2=1>0$. Since both are positive, the centers lie on the same side of the line. Thus, $x+2=0$ is a direct common tangent.

(B) $L_1: x - 2 = 0$. Distance from $O_1(2,3)$ to $L_1$: $d_1 = \frac{|2-2|}{\sqrt{1^2+0^2}} = 0 \neq r_1$. This is not a tangent.

(C) $L_1: 7x + 24y - 42 = 0$. Distance from $O_1(2,3)$ to $L_1$: $d_1 = \frac{|7(2) + 24(3) - 42|}{\sqrt{7^2+24^2}} = \frac{|14 + 72 - 42|}{\sqrt{49+576}} = \frac{|44|}{\sqrt{625}} = \frac{44}{25} \neq r_1$. This is not a tangent.

(D) $L_1: 7x - 24y + 42 = 0$. Distance from $O_1(2,3)$ to $L_1$: $d_1 = \frac{|7(2) - 24(3) + 42|}{\sqrt{7^2+(-24)^2}} = \frac{|14 - 72 + 42|}{\sqrt{49+576}} = \frac{|-16|}{\sqrt{625}} = \frac{16}{25} \neq r_1$. This is not a tangent.

Therefore, the equation of $L_1$ is $x+2=0$.