Question

The tangent to the circle \(C_1:x^2+y^2-2x-1=0\) at the point \((2,1)\) cuts off a chord of length \(4\) units from a circle \(C_2\) whose centre is \((3,-2)\). The radius of circle \(C_2\) is:

• A. \(\sqrt6\)
• B. \(2\)
• C. \(3\)
• D. \(2\sqrt2\)

Hint:

For a chord at perpendicular distance \(d\) from the center of a circle of radius \(r\): \[ \text{Chord length}=2\sqrt{r^2-d^2} \] Tangent at \((x_1,y_1)\) to circle \(x^2+y^2+2gx+2fy+c=0\): \[ xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0 \]

Answer 1

The Correct Option is A

Concept: The tangent at point \((x_1,y_1)\) to the circle: \[ x^2+y^2+2gx+2fy+c=0 \] is: \[ xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0 \] Also, if a chord of a circle of radius \(r\) is at perpendicular distance \(d\) from the center, then: \[ \text{Chord length}=2\sqrt{r^2-d^2} \]

Step 1: Find the tangent to \(C_1\). Given: \[ x^2+y^2-2x-1=0 \] Comparing with: \[ x^2+y^2+2gx+2fy+c=0 \] we get: \[ g=-1,\quad f=0,\quad c=-1 \] Point of tangency: \[ (2,1) \] Using tangent formula: \[ 2x+y-1(x+2)-1=0 \] \[ 2x+y-x-2-1=0 \] \[ x+y-3=0 \] Hence tangent is: \[ x+y-3=0 \]

Step 2: Find distance of center of \(C_2\) from this line. Center of \(C_2\): \[ (3,-2) \] Distance from line \(x+y-3=0\): \[ d=\frac{|3-2-3|}{\sqrt{1^2+1^2}} \] \[ d=\frac{|-2|}{\sqrt2} \] \[ d=\sqrt2 \]

Step 3: Use chord length formula. Chord length is: \[ 4 \] So: \[ 4=2\sqrt{r^2-d^2} \] \[ 2=\sqrt{r^2-2} \] Squaring: \[ 4=r^2-2 \] \[ r^2=6 \] \[ r=\sqrt6 \] Therefore the radius of \(C_2\) is: \[ \boxed{\sqrt6} \]