If the circles x^{2}+y^{2}-2x-2y-7=0 and x^{2 | vedclass

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

Question

(A) Given circles are $S_{1} \equiv x^{2}+y^{2}-2x-2y-7=0$ and $S_{2} \equiv x^{2}+y^{2}+4x+2y+k=0$.Here,$g_{1}=-1, f_{1}=-1, c_{1}=-7$ and $g_{2}=2,

Vedclass · Accepted Answer

$\frac{12}{\sqrt{13}}$

Vedclass · Answer

(A) Given circles are $S_{1} \equiv x^{2}+y^{2}-2x-2y-7=0$ and $S_{2} \equiv x^{2}+y^{2}+4x+2y+k=0$.Here,$g_{1}=-1, f_{1}=-1, c_{1}=-7$ and $g_{2}=2, f_{2}=1, c_{2}=k$.Since the circles cut orthogonally,$2(g_{1}g_{2} + f_{1}f_{2}) = c_{1} + c_{2}$.$2((-1)(2) + (-1)(1)) = -7 + k$$2(-2 - 1) = -7 + k$$-6 = -7 + k \Rightarrow k = 1$.The equation of the common chord is $S_{1} - S_{2} = 0$.$(x^{2}+y^{2}-2x-2y-7) - (x^{2}+y^{2}+4x+2y+1) = 0$$-6x - 4y - 8 = 0 \Rightarrow 3x + 2y + 4 = 0$.For circle $S_{1}$,center $C_{1} = (1, 1)$ and radius $r_{1} = \sqrt{1^{2} + 1^{2} - (-7)} = \sqrt{9} = 3$.The perpendicular distance $d$ from $C_{1}(1, 1)$ to the chord $3x + 2y + 4 = 0$ is:$d = \frac{|3(1) + 2(1) + 4|}{\sqrt{3^{2} + 2^{2}}} = \frac{|9|}{\sqrt{13}} = \frac{9}{\sqrt{13}}$.The length of the common chord is $2\sqrt{r_{1}^{2} - d^{2}}$.$= 2\sqrt{3^{2} - (\frac{9}{\sqrt{13}})^{2}} = 2\sqrt{9 - \frac{81}{13}} = 2\sqrt{\frac{117 - 81}{13}} = 2\sqrt{\frac{36}{13}} = \frac{12}{\sqrt{13}}$.

If the circles $x^{2}+y^{2}-2x-2y-7=0$ and $x^{2}+y^{2}+4x+2y+k=0$ cut orthogonally,then the length of the common chord of the circles is

Similar Questions

The combined equation of the direct common tangents of the circles $x^2+y^2-2x-2y-2=0$ and $x^2+y^2+4x+6y+12=0$ is:

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2-3x+y-10=0$ and $x^2+y^2-x+2y-20=0$ is

The length (in units) of the common chord of the circles $x^2+y^2+2x+3y+1=0$ and $x^2+y^2+4x+3y+2=0$ is:

The length of the chord of contact of the point $(2,1)$ with respect to the circle $x^2+y^2+4x+2y+1=0$ is

If the chord of contact of tangents from a point $A$ to a given circle passes through $B$,then the circle with $AB$ as a diameter will . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module