Respuesta :

[tex] {\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

Here we go ~

The general equation of ellipse is :

[tex]\qquad \sf  \dashrightarrow \: \cfrac{ {x}^{2} }{ {a}^{2} } + \cfrac{ {y}^{2} }{ {b}^{2} } = 1[/tex]

[ because it's a horizontal ellipse with centre at origin ]

so, let's equate given equation with the standard equation ~

[tex]\qquad \sf  \dashrightarrow \: \cfrac{ {x}^{2} }{64} + \cfrac{ {y}^{2} }{36} = 1[/tex]

se get :

  • a² = 64 ; a = 8

  • b² = 36 ; b = 6

As we know, length of minor axis is : 2b = 2 × 6 = 12 units

and, the smaller circle has diameter = 2b = 12 units

so, it's radius = 6 units ~

Now, let's write the equation of circle with origin as centre and radius = 6 units

[tex]\qquad \sf  \dashrightarrow \: {(x - h)}^{2} + (y - k) {}^{2} = {r}^{2} [/tex]

[ h = 0, k = 0, since circle has centre at origin ]

[tex]\qquad \sf  \dashrightarrow \: {x}^{2} + {y}^{2} = 36[/tex]

Answer:

[tex]x^2+y^2=36[/tex]

Step-by-step explanation:

Equation of the red eclipse (taken from your previously posted question):

[tex]\dfrac{x^2}{64}+\dfrac{y^2}{36}=1[/tex]

General equation of an ellipse

[tex]\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1[/tex]

where:

  • (h, k) is the center
  • [tex](h \pm a,k)\: \textsf{and}\:(h, k \pm b)\: \textsf{are the vertices}[/tex]

As the center of the red ellipse is (0, 0) ⇒ h = 0, k = 0.

[tex]\implies a^2=64 \implies a=\sqrt{64}=\pm8[/tex]

[tex]\implies b^2=36 \implies a=\sqrt{36}=\pm6[/tex]

The Major Axis is the longest diameter of an ellipse.

Therefore, using the formula for the vertices, the vertices of the major axis of the red ellipse are (8, 0) and (-8, 0).

The Minor Axis is the shortest diameter of an ellipse.

Therefore, using the formula for the vertices, the vertices of the minor axis of the red ellipse  (0, 6) and (0, -6).

Therefore, the radius of the larger circle is 8 units and the radius of the smaller circle is 6 units (half the respective axis, since the radius of a circle is half its diameter).

Equation of a circle

[tex](x-a)^2+(y-b)^2=r^2[/tex]

(where (a, b) is the center and r is the radius)

Given:

  • center = (0, 0)
  • radius = 6 units

The equation of the smaller circle is:

[tex]\implies (x-0)^2+(y-0)^2=6^2[/tex]

[tex]\implies x^2+y^2=36[/tex]

Learn more about equations of circles here:

https://brainly.com/question/27783217

https://brainly.com/question/27955619

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