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Q: The equation of the inner circle is x2 plus y2 equals 9 The radius of the outer circle is twice the radius of the inner circle What is the equation of the outer circle?

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The inner circle is x2 + y2 = 4. The radius of the inner circle is the square root of 4, which is 2. To find the radius of the outer circle, multiply 2 times 4. The radius of the outer circle is 8. Square 8 (82 or 8 x 8) to find the number to put into the equation of the outer circle. This is 64. The equation for the outer circle is x2 + y2 = 64.

The radius of the inner circle is sqrt(9) = 3. The radius of the outer circle is therefore 3*3 = 9. The equation for the outer circle is then x^2 + y^2 = 9^2 = 81. (This is an application of Pythagoras' theorem a^2 + b^2 = c^2 in a right-angled triangle by the way)

Draw two diameters perpendicular to each other. Draw a smaller circle with the same centre such that the radius of the inner circle is 'r' and the radius of the outer circle is 'r√2.' [Or, the radius of the outer circle is R and the radius of the inner circle is R/√2.]

it depends on what the radius is if they give you one

Continue or follow the inner or outer trajectory of the radius arc line in either, or both, directions until it meets itself. The i will have a center point which is at a fixed distance from the arc, equal to one half of the radius of the finished circle, and which will thereby prove the circle.

It's the radius of a circle or sphere WITH A SHELL(or Thickness); like if an Lid on a jar had a diameter of 11mm across on the outside (top of lid); and was ONE mm thick, then the inside radius would be 5. ie; 11mm outer diameter, minus (-) 1mm thickness (aka; or shell) equals (=) 10mm (inner) diameter; which is a 5mm radius.

From the Inner Circle was created in 2008.

Out of the Inner Circle was created in 1985.

A track 7m wide consists of two concentric circles. The circumference of the inner circle is 440m. Find, 1) the radius of the outer circle 2) the circumference of the outer circle.

Outer radius = a; inner radius = b; then area of the ring = pi (a2 - b2) = 90, where pi = 22/7. In view of a = 9, we have b2 = 81 - [90 x 7 / 22], the positive square root of which gives b.

As far as I know, there is no "regular" 4 pointed star. However, if you have a 4 pointed star, you can draw a circle through it's inner points and a second circle around its outer points. If we say that the inner circle has radius "r" and the outer circle has radius "R", that the angle, "a", of the star's points are: a = atan(2r / (R√2 - r) If the outer circle is twice as big as the inner circle, this becomes: a = atan(2r / (2r√2 - r) = atan(2 / (2√2-1) = 47.5°

The circles are concentric with centre (0,0). The radius of the outer circle is sqrt(72), that of the inner circle is sqrt(18). By Pythagoras, the length of the semichord is sqrt(72 - 18) = sqrt(54) units. Therefore the chord is 2*sqrt(54) = 6*sqrt(6) = 14.679 units (approx).

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