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

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

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.

Answer 1

The Correct Option is A

Step 1: Read the circle's centre and radius.
The circle is $x^2+y^2-2x+2y+c=0$, so $g=-1$, $f=1$. The centre is $(1,-1)$ and $r^2=g^2+f^2-c=1+1-c=2-c$.
Step 2: Find the distance from the centre to the chord line.
The chord lies on $5x-12y-4=0$. The distance from $(1,-1)$ is \[ d=\frac{|5(1)-12(-1)-4|}{\sqrt{25+144}}=\frac{|5+12-4|}{13}=\frac{13}{13}=1 \]
Step 3: Use the chord-length relation.
A chord of length $AB$ satisfies $AB^2=4(r^2-d^2)$, because half the chord, the radius, and $d$ form a right triangle. Here $AB=2\sqrt{3}$, so $AB^2=12$. \[ 12=4(r^2-1) \]
Step 4: Solve for $r^2$.
\[ 3=r^2-1\implies r^2=4 \]
Step 5: Find $c$.
Since $r^2=2-c$, we get $4=2-c$, so $c=-2$. The circle is $x^2+y^2-2x+2y-2=0$.
Step 6: Find the tangent length from $(2,1)$.
The tangent length is $\sqrt{S_{11}}$, where $S_{11}$ is the circle expression evaluated at the point: \[ S_{11}=2^2+1^2-2(2)+2(1)-2=4+1-4+2-2=1 \] \[ \text{Tangent length}=\sqrt{1}=\boxed{1} \]