The number of common tangents to the ellipse | vedclass

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(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

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$1$

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(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:

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