The angle of intersection of the circles x^2 | vedclass

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

Question

(B) For the circle $x^2 + y^2 + 8x - 2y - 9 = 0$,the center $C_1 = (-4, 1)$ and radius $r_1 = \sqrt{(-4)^2 + 1^2 - (-9)} = \sqrt{16 + 1 + 9} = \sqrt{2

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(B) For the circle $x^2 + y^2 + 8x - 2y - 9 = 0$,the center $C_1 = (-4, 1)$ and radius $r_1 = \sqrt{(-4)^2 + 1^2 - (-9)} = \sqrt{16 + 1 + 9} = \sqrt{26}$.For the circle $x^2 + y^2 - 2x + 8y - 7 = 0$,the center $C_2 = (1, -4)$ and radius $r_2 = \sqrt{1^2 + (-4)^2 - (-7)} = \sqrt{1 + 16 + 7} = \sqrt{24}$.The distance between the centers $d^2 = (C_1C_2)^2 = (1 - (-4))^2 + (-4 - 1)^2 = 5^2 + (-5)^2 = 25 + 25 = 50$.The angle of intersection $	heta$ is given by $\cos 	heta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2}$.Substituting the values: $\cos 	heta = \frac{26 + 24 - 50}{2 \sqrt{26} \sqrt{24}} = \frac{50 - 50}{2 \sqrt{624}} = 0$.Since $\cos 	heta = 0$,we have $	heta = 90^o$.

The angle of intersection of the circles $x^2 + y^2 + 8x - 2y - 9 = 0$ and $x^2 + y^2 - 2x + 8y - 7 = 0$ is ............ $^o$.

Similar Questions

Let the circle with centre at origin pass through the vertices of an equilateral triangle $ABC$. If $A = (2, 4)$,then the length of the median through $A$ is

The intercept on the line $y=x$ by the circle $x^2+y^2-2x=0$ is $AB$. The equation of the circle with $AB$ as diameter is:

If $\sin ^{-1} a$ is the acute angle between the curves $x^{2}+y^{2}=4 x$ and $x^{2}+y^{2}=8$ at $(2,2)$,then $a$ is equal to

If the angle between the circles $x^2+y^2+4x-5=0$ and $x^2+y^2+2\lambda y-4=0$ is $\frac{\pi}{3}$,then $\lambda=$

If $\sin ^{-1}(a)$ is the acute angle between the curves $x^2+y^2=4x$ and $x^2+y^2=8$ at the point $(2,2)$,then $a$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module