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

The locus of a point which moves such that the sum of the squares of its distances from the sides of a square of side length $1$ is $9$,is

$A$ point moves in such a way that its distance from the origin is always $4$. Then the locus of the point is:

If $A (c, 0)$ and $B (-c, 0)$ are two points,find the locus of point $P$ such that $PA^{2} + PB^{2} = AB^{2}$.

$A$ point $P(x, y)$ moves such that the sum of the squares of its distances from the points $(1, 2)$ and $(-2, 1)$ is $14$. Let $f(x, y) = 0$ be the locus of $P$,which intersects the $x$-axis at points $A, B$ and the $y$-axis at points $C, D$. Then the area of the quadrilateral $ACBD$ is equal to:

The set of all real values of $\lambda$ for which the point $P$ with coordinates $(\lambda, \lambda^2)$ does not lie inside the triangle formed by the lines $x - y = 0$, $x + y - 2 = 0$, and $x + 3 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module