Let A be the point (3, 0) and circles with va | vedclass

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

Question

(B) Let $B = (x, y)$. The diameter of the circle is $AB$, where $A = (3, 0)$.The center of the circle is $M = (\frac{x+3}{2}, \frac{y}{2})$ and its ra

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) Let $B = (x, y)$. The diameter of the circle is $AB$, where $A = (3, 0)$.The center of the circle is $M = (\frac{x+3}{2}, \frac{y}{2})$ and its radius is $r = \frac{1}{2} \sqrt{(x-3)^2 + y^2}$.The circle $x^2 + y^2 = 36$ has center $O = (0, 0)$ and radius $R = 6$.Since the circles touch internally, the distance between centers $OM = R - r$.$OM = \sqrt{(\frac{x+3}{2})^2 + (\frac{y}{2})^2} = 6 - \frac{1}{2} \sqrt{(x-3)^2 + y^2}$.Multiplying by $2$: $\sqrt{(x+3)^2 + y^2} = 12 - \sqrt{(x-3)^2 + y^2}$.Let $d_1 = \sqrt{(x-3)^2 + y^2}$ (distance $AB$) and $d_2 = \sqrt{(x+3)^2 + y^2}$ (distance $OB$).The equation is $d_2 = 12 - d_1$, or $d_1 + d_2 = 12$.Since $d_1 + d_2 = 12 > AB = 6$, the locus of $B$ is an ellipse with foci at $A(3, 0)$ and $O(-3, 0)$.The distance between foci $2ae = 6$, so $ae = 3$.The major axis $2a = 12$, so $a = 6$.Thus, $e = \frac{3}{6} = \frac{1}{2}$.Then $e^2 = \frac{1}{4}$.Finally, $72e^2 = 72 	imes \frac{1}{4} = 18$.

Let $A$ be the point $(3, 0)$ and circles with variable diameter $AB$ touch the circle $x^2 + y^2 = 36$ internally. Let the curve $C$ be the locus of the point $B$. If the eccentricity of $C$ is $e$, then $72e^2$ is equal to . . . . . .

Similar Questions

If the sum of the distances of a point from the origin and the line $x = 2$ is $4$,then its locus is

If $P(x_1, y_1)$ is a point such that the lengths of the tangents from it to the circles $x^2+y^2-4x-6y-12=0$ and $x^2+y^2+6x+18y+26=0$ are in the ratio $2:3$,then the locus of $P$ is

The locus of the point of intersection of the tangents to the circle $x^2+y^2=a^2$ which make complementary angles with the $X$-axis is

The locus of the mid-points of the chords of the circle $x^2 + y^2 - 2x - 4y - 11 = 0$ which subtend $60^o$ at the centre is

If $A$ and $B$ are the points of intersection of the circle $x^2+y^2-8x=0$ and the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$,and a point $P$ moves on the line $2x-3y+4=0$,then the centroid of $\triangle PAB$ lies on the line:

Vedclass Test Series

Exam Paper Generator

Online Exam Module