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Q: What is the general form of the equation for the given circle centered at o (00)?

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Exactly as it's stated, that equation describes a straight line, not a circle. If you take out the phrase "times 2" from both places where it's used and replace it with "squared", then the equation describes a circle, centered at (-5, 3), with a radius of 5.

It's at the point ( -5, 3 ) The general formula of a circle is given by the equation: ( x - h )2 + ( y - k )2 = r2 Where the point ( h, k ) is the center of the circle, and r is the radius of the circle. In the equation ( x + 5 )2 + ( y - 3 )2 = 25 The first group of terms isn't subtracted. Rather than x + 5, this is actually x - -5. The center of the circle, then, is the point ( -5, 3 )

The general equation to represent a line isaX + bY = c, where a, b, and c are given values or parameters.See the reference for more information.

Square feet are a measure of area, and the area of a circle is usually given by the equation: a = πr2 The circumference of a circle is given with the equation: c = 2πr To find the area of the circle given only it's circumference, we can rearrange the circumference equation, solving it for r, and then substituting it into our radius equation in place of r. c = 2πr r = c / 2π a = πr2 a = π(c / 2π)2 a = πc2 / 4π2 a = c2/π So the area of a circle is equal to the square of it's circumference divided by pi.

What is the center of the circle given by the equation (x- 2)2 + (y + 4)2 = 6?(2, -4)

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The general equation of a circle is given by the formula(x - h)2 + (x - k)2 = r2, where (h, k) is the center of the circle, and r its radius.Since the center of the circle is (0, 0), the equation reduces tox2 + y2 = r2So that the equation of our circle is x2 + y2 = 36.

a circle, centered at the given point.

Exactly as it's stated, that equation describes a straight line, not a circle. If you take out the phrase "times 2" from both places where it's used and replace it with "squared", then the equation describes a circle, centered at (-5, 3), with a radius of 5.

You cannot show it in general since it need not be true!

It's at the point ( -5, 3 ) The general formula of a circle is given by the equation: ( x - h )2 + ( y - k )2 = r2 Where the point ( h, k ) is the center of the circle, and r is the radius of the circle. In the equation ( x + 5 )2 + ( y - 3 )2 = 25 The first group of terms isn't subtracted. Rather than x + 5, this is actually x - -5. The center of the circle, then, is the point ( -5, 3 )

The center of the circle given by the equation (x - 3)2 plus (y + 2)2 = 9 is (3,-2).

The general equation to represent a line isaX + bY = c, where a, b, and c are given values or parameters.See the reference for more information.

Square feet are a measure of area, and the area of a circle is usually given by the equation: a = πr2 The circumference of a circle is given with the equation: c = 2πr To find the area of the circle given only it's circumference, we can rearrange the circumference equation, solving it for r, and then substituting it into our radius equation in place of r. c = 2πr r = c / 2π a = πr2 a = π(c / 2π)2 a = πc2 / 4π2 a = c2/π So the area of a circle is equal to the square of it's circumference divided by pi.

What is the center of the circle given by the equation (x- 2)2 + (y + 4)2 = 6?(2, -4)

It's a circle. The equation of a circle is x^2+y^2=r^2. So the equation you've given is a circle with a radius of 4 and, since there are no modifications to the x or y values, the center of the circle is located at (0,0).

multiple the diameter by 2 to get the radius, then use your equation and go from there

the general equation of the circle is (x-a)^2+(y-b)^2=r^2 where (a,b) is the center of the circle. r is the radius of the circle substituting the given values, (x+4)^2+(y-3)^2=6^2 x^2+y^2+8x-6y+16+9=36 x^2+y^2+8x-6y=11