Question

If the length of the chord 2x+3y+k=0 of the circle \(x^2+y^2-2x+4y-11=0\) is \(2\sqrt{3}\), then the sum of all possible values of k is

• A. 26
• B. 8
• C. 13
• D. 4

Hint:

A chord, the radius to one of its endpoints, and the perpendicular from the center form a right-angled triangle. The hypotenuse is the radius (r), one leg is the perpendicular distance (d), and the other leg is half the chord length (L/2). This gives the relation \(r^2 = d^2 + (L/2)^2\), which is fundamental for chord problems.

Answer 1

The Correct Option is B

Step 1: Find the center and radius of the given circle.
The equation is \(x^2+y^2-2x+4y-11=0\).
The center \(C\) is \((-g, -f) = (1, -2)\).
The radius \(r\) is \( \sqrt{g^2+f^2-c} = \sqrt{1^2 + (-2)^2 - (-11)} = \sqrt{1+4+11} = \sqrt{16} = 4 \).
Step 2: Use the formula for the length of a chord.
The length of the chord \(L\) is given by \( L = 2\sqrt{r^2 - d^2} \), where \(d\) is the perpendicular distance from the center to the chord.
We are given \(L = 2\sqrt{3}\).
\( 2\sqrt{3} = 2\sqrt{4^2 - d^2} \implies \sqrt{3} = \sqrt{16 - d^2} \).
Squaring both sides: \( 3 = 16 - d^2 \implies d^2 = 13 \implies d = \sqrt{13} \).
Step 3: Calculate the distance d and solve for k.
The distance from the center \((1, -2)\) to the line \(2x+3y+k=0\) is given by the formula:
\( d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2+B^2}} = \frac{|2(1) + 3(-2) + k|}{\sqrt{2^2+3^2}} = \frac{|2-6+k|}{\sqrt{13}} = \frac{|k-4|}{\sqrt{13}} \).
We know \(d = \sqrt{13}\), so \( \frac{|k-4|}{\sqrt{13}} = \sqrt{13} \).
\( |k-4| = 13 \).
This gives two possible equations:
1) \( k-4 = 13 \implies k_1 = 17 \).
2) \( k-4 = -13 \implies k_2 = -9 \).
Step 4: Find the sum of the possible values of k.
Sum = \( k_1 + k_2 = 17 + (-9) = 8 \).