Question:
The line \( 5x-12y-4=0 \) cuts the circle \( x^{2}+y^{2}-2x+2y+c=0 \) at two points A, B. If \( AB=2\sqrt{3} \), then the length of the tangent drawn from the point (2, 1) to the given circle is
The line \( 5x-12y-4=0 \) cuts the circle \( x^{2}+y^{2}-2x+2y+c=0 \) at two points A, B. If \( AB=2\sqrt{3} \), then the length of the tangent drawn from the point (2, 1) to the given circle is
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Always use \(AB^2 = 4(r^2 - d^2)\) instead of computing intersection points directly—it saves a lot of algebra in chord-based circle problems.
Updated On: Jun 8, 2026
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The Correct Option is A
Solution and Explanation
Concept:
• For a circle \(x^2+y^2+2gx+2fy+c=0\), center is \((-g,-f)\) and \(r^2=g^2+f^2-c\).
• Chord length relation: \(AB^2 = 4(r^2 - d^2)\), where \(d\) is distance of center from chord.
• Tangent length from point \(P\): \(PT = \sqrt{S_{11}}\).
Step 1: Identify center and radius expression.
Given circle: \[ x^2 + y^2 - 2x + 2y + c = 0 \] So, \[ g = -1,\quad f = 1,\quad \text{center } C(1,-1),\quad r^2 = 1+1-c = 2-c \]
Step 2: Find radius using chord length formula.
Given line: \[ 5x - 12y - 4 = 0 \] Distance of center from line: \[ d = \frac{|5(1) - 12(-1) - 4|}{\sqrt{25+144}} = \frac{|5+12-4|}{13} = \frac{13}{13} = 1 \] Chord length: \[ AB = 2\sqrt{3} \Rightarrow AB^2 = 12 \] Using: \[ AB^2 = 4(r^2 - d^2) \] \[ 12 = 4(r^2 - 1) \Rightarrow 3 = r^2 - 1 \Rightarrow r^2 = 4 \]
Step 3: Find constant \(c\).
\[ r^2 = 2 - c \Rightarrow 4 = 2 - c \Rightarrow c = -2 \]
Step 4: Length of tangent from (2,1).
\[ S_{11} = x_1^2 + y_1^2 - 2x_1 + 2y_1 + c \] Substitute \( (2,1) \): \[ S_{11} = 4 + 1 - 4 + 2 - 2 = 1 \] \[ PT = \sqrt{S_{11}} = 1 \] \[ \boxed{1} \]
• For a circle \(x^2+y^2+2gx+2fy+c=0\), center is \((-g,-f)\) and \(r^2=g^2+f^2-c\).
• Chord length relation: \(AB^2 = 4(r^2 - d^2)\), where \(d\) is distance of center from chord.
• Tangent length from point \(P\): \(PT = \sqrt{S_{11}}\).
Step 1: Identify center and radius expression.
Given circle: \[ x^2 + y^2 - 2x + 2y + c = 0 \] So, \[ g = -1,\quad f = 1,\quad \text{center } C(1,-1),\quad r^2 = 1+1-c = 2-c \]
Step 2: Find radius using chord length formula.
Given line: \[ 5x - 12y - 4 = 0 \] Distance of center from line: \[ d = \frac{|5(1) - 12(-1) - 4|}{\sqrt{25+144}} = \frac{|5+12-4|}{13} = \frac{13}{13} = 1 \] Chord length: \[ AB = 2\sqrt{3} \Rightarrow AB^2 = 12 \] Using: \[ AB^2 = 4(r^2 - d^2) \] \[ 12 = 4(r^2 - 1) \Rightarrow 3 = r^2 - 1 \Rightarrow r^2 = 4 \]
Step 3: Find constant \(c\).
\[ r^2 = 2 - c \Rightarrow 4 = 2 - c \Rightarrow c = -2 \]
Step 4: Length of tangent from (2,1).
\[ S_{11} = x_1^2 + y_1^2 - 2x_1 + 2y_1 + c \] Substitute \( (2,1) \): \[ S_{11} = 4 + 1 - 4 + 2 - 2 = 1 \] \[ PT = \sqrt{S_{11}} = 1 \] \[ \boxed{1} \]
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