Question

The straight line touching the circle \(x^2+y^2-2x-3=0\) and remaining normal to the circle \(x^2+y^2-4y-6=0\) is

• A. \( 4x-3y+6=0 \)
• B. \( y+2=0 \)
• C. \( 4x+3y-6=0 \)
• D. \( 2x+3=0 \)

Hint:

Center of circle \(x^2+y^2+2gx+2fy+c=0\) is \((-g,-f)\), radius is \(\sqrt{g^2+f^2-c}\).
A line normal to a circle passes through its center.
A line tangent to a circle has its perpendicular distance from the center equal to the radius.
Distance from point \((x_0,y_0)\) to line \(Ax+By+C=0\) is \(|Ax_0+By_0+C|/\sqrt{A^2+B^2}\).

Answer 1

The Correct Option is A

Let Circle 1 (C1) be \(x^2+y^2-2x-3=0\). Center \(C_1 = (-g, -f)\). Here \(2g=-2 \Rightarrow g=-1\); \(2f=0 \Rightarrow f=0\). So \(C_1 = (1,0)\). Radius \(r_1 = \sqrt{g^2+f^2-c} = \sqrt{(-1)^2+0^2-(-3)} = \sqrt{1+3} = \sqrt{4} = 2\). Let Circle 2 (C2) be \(x^2+y^2-4y-6=0\). Center \(C_2\). Here \(2g=0 \Rightarrow g=0\); \(2f=-4 \Rightarrow f=-2\). So \(C_2 = (0,2)\). Radius \(r_2 = \sqrt{0^2+(-2)^2-(-6)} = \sqrt{4+6} = \sqrt{10}\). Let the required straight line be L. L is tangent to C1. L is normal to C2. A line normal to a circle passes through its center. So, line L passes through the center of C2, which is \(C_2(0,2)\). Since L is tangent to C1, the perpendicular distance from the center of C1 (\(C_1(1,0)\)) to line L must be equal to the radius of C1 (\(r_1=2\)). Let the equation of line L be \(y-2 = m(x-0) \Rightarrow y-2 = mx \Rightarrow mx - y + 2 = 0\). (Line L passes through \(C_2(0,2)\) with slope m). The distance from \(C_1(1,0)\) to line \(mx-y+2=0\) is \(r_1=2\). Distance formula: \(d = \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}\). Here, \((x_0,y_0)=(1,0)\), line is \(mx-1y+2=0\). So \(A=m, B=-1, C=2\). \[ 2 = \frac{|m(1) - 1(0) + 2|}{\sqrt{m^2+(-1)^2}} = \frac{|m+2|}{\sqrt{m^2+1}} \] Square both sides: \(4 = \frac{(m+2)^2}{m^2+1}\) \(4(m^2+1) = (m+2)^2\) \(4m^2+4 = m^2+4m+4\) \(3m^2 - 4m = 0\) \(m(3m-4) = 0\). So, \(m=0\) or \(3m-4=0 \Rightarrow m=4/3\). Case 1: \(m=0\). Line L is \(0x - y + 2 = 0 \Rightarrow -y+2=0 \Rightarrow y-2=0\) or \(y=2\). This is option (b) if written as \(y+2=0\). Wait, \(y-2=0\) is \(y=2\). Option (b) is \(y+2=0\), which is \(y=-2\). So \(y=2\) is not matching (b). The question is \(y+2=0\). My solution is \(y-2=0\). Case 2: \(m=4/3\). Line L is \((4/3)x - y + 2 = 0\). Multiply by 3: \(4x - 3y + 6 = 0\). This matches option (a). Let's check option (b) \(y+2=0\). This means \(y=-2\). This line passes through \(C_2(0,2)\) only if \(2=-2\), which is false. So slope method is better. Line \(y+2=0\) is \(0x+y+2=0\). Distance from \(C_1(1,0)\) is \(\frac{|0(1)+1(0)+2|}{\sqrt{0^2+1^2}} = \frac{2}{1}=2\). So it is tangent to C1. Does \(y+2=0\) pass through \(C_2(0,2)\)? Substitute \((0,2)\): \(2+2=4 \neq 0\). No. So option (b) is incorrect. My derived lines are \(y-2=0\) (i.e., \(y=2\)) and \(4x-3y+6=0\). Option (a) is \(4x-3y+6=0\). This is one of my solutions. The checkmark in image is on option (a). \[ \boxed{4x-3y+6=0} \]