The number of common tangents to the ellipse | vedclass

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

Question

(B) The equation of the ellipse is $\frac{(x - 2)^2}{3^2} + \frac{(y + 2)^2}{2^2} = 1$. Its center is $(2, -2)$ and semi-axes are $a=3, b=2$.The equat

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The equation of the ellipse is $\frac{(x - 2)^2}{3^2} + \frac{(y + 2)^2}{2^2} = 1$. Its center is $(2, -2)$ and semi-axes are $a=3, b=2$.The equation of the circle is $x^2 + y^2 - 4x + 2y + 4 = 0$. Completing the square: $(x-2)^2 + (y+1)^2 = 1$. Its center is $(2, -1)$ and radius $r=1$.Let the center of the ellipse be $C_1(2, -2)$ and the center of the circle be $C_2(2, -1)$. The distance between centers $d = \sqrt{(2-2)^2 + (-1 - (-2))^2} = 1$.For the ellipse,the vertical distance from the center $(2, -2)$ to the top vertex is $b=2$. Since the circle lies within the vertical range of the ellipse and the distance between centers is small,we check the intersection. Substituting $x=2$ into the ellipse equation gives $\frac{0}{9} + \frac{(y+2)^2}{4} = 1$,so $(y+2)^2 = 4$,$y = 0$ or $y = -4$. The circle at $x=2$ gives $(y+1)^2 = 1$,so $y=0$ or $y=-2$. Both curves pass through $(2, 0)$.Since the circle is contained within the ellipse and they touch at $(2, 0)$,there is exactly $1$ common tangent at the point of contact.

The number of common tangents to the ellipse $\frac{(x - 2)^2}{9} + \frac{(y + 2)^2}{4} = 1$ and the circle $x^2 + y^2 - 4x + 2y + 4 = 0$ is:

Similar Questions

Among the chords of the circle $x^2+y^2=75$,the number of chords having their midpoints on the line $x=8$ and having their slopes as integers is

The sum of the squares of the lengths of the chords intercepted on the circle,$x^2 + y^2 = 16$,by the lines,$x + y = n$,$n \in N$,where $N$ is the set of all natural numbers,is

Let $C$ be a circle passing through the points $A (2,-1)$ and $B (3,4)$. The line segment $AB$ is not a diameter of $C$. If $r$ is the radius of $C$ and its centre lies on the circle $(x-5)^{2}+(y-1)^{2}=\frac{13}{2}$,then $r^{2}$ is equal to

The circles $x^2 + y^2 - 10x + 16 = 0$ and $x^2 + y^2 = r^2$ intersect each other in two distinct points,if

$A$ square is inscribed in the circle $x^2 + y^2 - 2x + 4y + 3 = 0$,whose sides are parallel to the coordinate axes. One vertex of the square is

Vedclass Test Series

Exam Paper Generator

Online Exam Module