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

Step 1: Read off circle $C_1$.
$C_1:x^2+y^2-2x-1=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$ gives $g=-1$, $f=0$, $c=-1$.
Step 2: Write the tangent at $(2,1)$.
The tangent at $(x_1,y_1)$ is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$. Putting the values: \[ 2x+y-1(x+2)-1=0. \]
Step 3: Simplify the tangent.
$2x+y-x-2-1=0$, so the tangent line is $x+y-3=0$.
Step 4: Find the distance from the centre of $C_2$.
Centre of $C_2$ is $(3,-2)$. Distance to $x+y-3=0$: \[ d=\frac{|3-2-3|}{\sqrt{1^2+1^2}}=\frac{2}{\sqrt2}=\sqrt2. \]
Step 5: Use the chord length rule.
The chord cut from $C_2$ has length $4$, and chord $=2\sqrt{r^2-d^2}$: \[ 4=2\sqrt{r^2-2}\Rightarrow 2=\sqrt{r^2-2}. \]
Step 6: Solve for $r$.
Square: $4=r^2-2$, so $r^2=6$ and $r=\sqrt6$. \[ \boxed{\sqrt6} \]