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
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
Show Hint
A tangent to a circle is at a distance equal to the radius from the centre, while a normal to a circle must pass through the centre.
Updated On: Jun 26, 2026
- \(4x-3y+6=0\)
- \(y+2=0\)
- \(4x+3y-6=0\)
- \(2x+3=0\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Find the centre and radius of the first circle.
Given, \[ x^2+y^2-2x-3=0 \] Completing the square, \[ (x-1)^2+y^2=4 \] So, the centre is \[ C_1=(1,0) \] and radius is \[ r_1=2. \]
Step 2: Find the centre of the second circle.
Given, \[ x^2+y^2-4y-6=0 \] Completing the square, \[ x^2+(y-2)^2=10 \] So, the centre is \[ C_2=(0,2). \] A normal to a circle always passes through its centre.
Step 3: Check the given line.
For option (1), \[ 4x-3y+6=0. \] Substitute \(C_2=(0,2)\): \[ 4(0)-3(2)+6=0 \] \[ -6+6=0 \] So, the line passes through the centre of the second circle. Hence, it is normal to the second circle.
Step 4: Check whether the line is tangent to the first circle.
Distance of \(C_1=(1,0)\) from the line \[ 4x-3y+6=0 \] is \[ d=\frac{|4(1)-3(0)+6|}{\sqrt{4^2+(-3)^2}} \] \[ d=\frac{10}{5}=2 \] Since \[ d=r_1=2, \] the line touches the first circle.
Step 5: Final conclusion.
Therefore, \[ \boxed{4x-3y+6=0} \]
Given, \[ x^2+y^2-2x-3=0 \] Completing the square, \[ (x-1)^2+y^2=4 \] So, the centre is \[ C_1=(1,0) \] and radius is \[ r_1=2. \]
Step 2: Find the centre of the second circle.
Given, \[ x^2+y^2-4y-6=0 \] Completing the square, \[ x^2+(y-2)^2=10 \] So, the centre is \[ C_2=(0,2). \] A normal to a circle always passes through its centre.
Step 3: Check the given line.
For option (1), \[ 4x-3y+6=0. \] Substitute \(C_2=(0,2)\): \[ 4(0)-3(2)+6=0 \] \[ -6+6=0 \] So, the line passes through the centre of the second circle. Hence, it is normal to the second circle.
Step 4: Check whether the line is tangent to the first circle.
Distance of \(C_1=(1,0)\) from the line \[ 4x-3y+6=0 \] is \[ d=\frac{|4(1)-3(0)+6|}{\sqrt{4^2+(-3)^2}} \] \[ d=\frac{10}{5}=2 \] Since \[ d=r_1=2, \] the line touches the first circle.
Step 5: Final conclusion.
Therefore, \[ \boxed{4x-3y+6=0} \]
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