If the length of the chord 2x + 3y + k = 0 of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given circle is $x^2 + y^2 - 2x + 4y - 11 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$,$f = 2$,and $c = -11$. The cent

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) The given circle is $x^2 + y^2 - 2x + 4y - 11 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -1$,$f = 2$,and $c = -11$. The center of the circle is $(-g, -f) = (1, -2)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + 2^2 - (-11)} = \sqrt{1 + 4 + 11} = \sqrt{16} = 4$. The length of the chord is $L = 2\sqrt{3}$. Let $d$ be the perpendicular distance from the center $(1, -2)$ to the chord $2x + 3y + k = 0$. Using the formula $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$,we have $d = \frac{|2(1) + 3(-2) + k|}{\sqrt{2^2 + 3^2}} = \frac{|2 - 6 + k|}{\sqrt{13}} = \frac{|k - 4|}{\sqrt{13}}$. In a circle,$r^2 = d^2 + (L/2)^2$. Substituting the values,$4^2 = d^2 + (\sqrt{3})^2$,which gives $16 = d^2 + 3$,so $d^2 = 13$. Thus,$d = \sqrt{13}$. Equating the two expressions for $d$,$\frac{|k - 4|}{\sqrt{13}} = \sqrt{13}$,which implies $|k - 4| = 13$. This gives $k - 4 = 13$ or $k - 4 = -13$,so $k = 17$ or $k = -9$. The sum of all possible values of $k$ is $17 + (-9) = 8$.

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

Similar Questions

The radius of a circle $C_1$ is thrice the radius of another circle $C_2$. The centres of $C_1$ and $C_2$ are $(1, 2)$ and $(3, -2)$ respectively. If they cut each other orthogonally,find the equation of the circle with radius $r$ and centre $(1, -2)$.

For the circle $x^2 + y^2 + 6x - 8y + 9 = 0$,which of the following statements is true?

The radius of the circle passing through the points $(5,7)$,$(2,-2)$,and $(-2,0)$ is (in $units$)

The number of circles that touch all the straight lines $x+y=4$,$x-y=-2$,and $y=2$ is

The length of the intercept that the circle $x^2 + y^2 + 10x - 6y + 9 = 0$ makes on the $x$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module