If the angle between the circles x^2+y^2-2x-4 | vedclass

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

Question

(C) The equations of the circles are $S_1: x^2+y^2-2x-4y+c=0$ and $S_2: x^2+y^2-4x-2y+4=0$. For $S_1$,the center $C_1 = (1, 2)$ and radius $r_1 = \sqr

Vedclass · Accepted Answer

$\frac{7 \pm \sqrt{5}}{2}$

Vedclass · Answer

(C) The equations of the circles are $S_1: x^2+y^2-2x-4y+c=0$ and $S_2: x^2+y^2-4x-2y+4=0$. For $S_1$,the center $C_1 = (1, 2)$ and radius $r_1 = \sqrt{1^2+2^2-c} = \sqrt{5-c}$. For $S_2$,the center $C_2 = (2, 1)$ and radius $r_2 = \sqrt{2^2+1^2-4} = 1$. The distance between the centers $d = \sqrt{(2-1)^2 + (1-2)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$. The angle $	heta$ between two circles is given by $\cos 	heta = \left| \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} ight|$. Given $	heta = 60^{\circ}$,so $\cos 60^{\circ} = \frac{1}{2}$. Substituting the values: $\frac{1}{2} = \left| \frac{(5-c) + 1 - 2}{2 \cdot \sqrt{5-c} \cdot 1} ight| = \left| \frac{4-c}{2\sqrt{5-c}} ight|$. Squaring both sides: $\frac{1}{4} = \frac{(4-c)^2}{4(5-c)} \Rightarrow 5-c = (4-c)^2$. $5-c = 16 - 8c + c^2 \Rightarrow c^2 - 7c + 11 = 0$. Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $c = \frac{7 \pm \sqrt{49-44}}{2} = \frac{7 \pm \sqrt{5}}{2}$.

If the angle between the circles $x^2+y^2-2x-4y+c=0$ and $x^2+y^2-4x-2y+4=0$ is $60^{\circ}$,then $c=$

Similar Questions

$A$ line is drawn through a fixed point $P(\alpha, \beta)$ to cut the circle $x^{2}+y^{2}=r^{2}$ at $A$ and $B$. Then $PA \cdot PB$ is equal to

Two circles $x^2 + y^2 = ax$ and $x^2 + y^2 = c^2$ touch each other if:

If the coordinates of one end of the diameter of the circle $x^2 + y^2 - 8x - 4y + c = 0$ are $(-3, 2)$,then the coordinates of the other end are:

The line $y=mx+c$ intercepts the circle $x^2+y^2=r^2$ in two distinct points,if

$A$ triangle $PQR$ is inscribed in the circle $x^2 + y^2 = 25$. If the coordinates of $Q$ and $R$ are $(3, 4)$ and $(-4, 3)$ respectively,then $\angle QPR = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module