Question

If the pole of the line \[ 2x+3y-20=0 \] with respect to the circle \[ x^2+y^2-4x+6y-12=0 \] is \((\alpha,\beta)\), then the number of tangents that can be drawn from \((\alpha,\beta)\) to the given circle is:

• A. \(\beta\)
• B. \(\alpha-2\)
• C. \(\beta+1\)
• D. \(\alpha\)

Hint:

For any point with distance \(d\) from the centre of a circle of radius \(r\): \[ d>r \Rightarrow 2 \text{ tangents} \] \[ d=r \Rightarrow 1 \text{ tangent} \] \[ d<r \Rightarrow 0 \text{ tangents} \] This test quickly determines the number of tangents from a point to a circle.

Answer 1

The Correct Option is A

Concept: For the circle \[ x^2+y^2+2gx+2fy+c=0, \] the polar of a point \((x_1,y_1)\) is given by \[ xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0. \] In this problem, the given line is the polar of the point \((\alpha,\beta)\). Therefore, by comparing the equation of the polar with the given line, we can determine the coordinates of the pole. Once the pole is obtained, we locate the point with respect to the circle. If the point lies:

• Outside the circle \(\Rightarrow\) two tangents can be drawn.

• On the circle \(\Rightarrow\) one tangent can be drawn.

• Inside the circle \(\Rightarrow\) no tangent can be drawn.
Thus, after finding \((\alpha,\beta)\), we simply check whether the point lies inside, on, or outside the circle.

Step 1: Write the circle in standard form. The given circle is \[ x^2+y^2-4x+6y-12=0. \] Comparing with \[ x^2+y^2+2gx+2fy+c=0, \] we obtain \[ g=-2,\qquad f=3,\qquad c=-12. \]

Step 2: Write the polar of the point \((\alpha,\beta)\). Using the polar formula, \[ x\alpha+y\beta-2(x+\alpha)+3(y+\beta)-12=0. \] Expanding, \[ \alpha x+\beta y-2x-2\alpha+3y+3\beta-12=0. \] Grouping coefficients, \[ (\alpha-2)x+(\beta+3)y+(-2\alpha+3\beta-12)=0. \]

Step 3: Compare with the given line. The given line is \[ 2x+3y-20=0. \] Comparing coefficients, \[ \alpha-2=2, \] which gives \[ \alpha=4. \] Also, \[ \beta+3=3, \] which gives \[ \beta=0. \] Checking the constant term, \[ -2(4)+3(0)-12=-20, \] which agrees perfectly with the given line. Therefore, \[ (\alpha,\beta)=(4,0). \]

Step 4: Find the centre and radius of the circle. Completing squares, \[ x^2-4x+y^2+6y-12=0, \] \[ (x-2)^2-4+(y+3)^2-9-12=0, \] \[ (x-2)^2+(y+3)^2=25. \] Hence, \[ \text{Centre }=(2,-3), \] and \[ r=5. \]

Step 5: Find the distance of the pole from the centre. Distance between \((4,0)\) and \((2,-3)\) is \[ d=\sqrt{(4-2)^2+(0+3)^2}. \] \[ =\sqrt{4+9}. \] \[ =\sqrt{13}. \] Since \[ \sqrt{13}<5, \] the point \((4,0)\) lies

inside the circle.

Step 6: Determine the number of tangents. A point lying inside a circle cannot have any real tangent drawn to the circle. Therefore, \[ \text{Number of tangents}=0. \] Since \[ \beta=0, \] the number of tangents is equal to \[ \boxed{\beta}. \]

Step 7: Final Conclusion. The required number of tangents is \[ 0, \] which is represented by \[ \boxed{\beta}. \] Hence the correct answer is \[ \boxed{\text{Option (A)}}. \]