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Q: Where is the center of the circle given by the equation x - 3 squared plus y plus 2 squared equals 9?

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(-4,3)

(-4,-6)

You are describing a circle, with its center at the origin and a radius of 4 (the square root of 16)

The centre is (-5, 3)

I think this means (x+7)squared + (y-5)squared =4squared, which represents a circle centred at -7,+5

The equation describes a circle with its centre at the origin and radius = âˆš13. Each and every point on that circle is a solution.

Area of a circle equals pi (~3.14) multiplied by the radius squared. So to find the area of that circle your equation is (3.14)(3)^2 which equals approximately 27.

x2 + y2 = 49

no

No. C = 2*pi*r is the equation representing the circumference of a circle. The area of a circle is equal to pi*(r^2).

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.

The area of a circle is defined by the equation A = πr^2 where π = pi (3.14...) A = Area r^2 = radius squared So, the radius of a circle squared multiplied by pi equals the area.

56

Divide all terms by 2 and complete the square for x and y:- So: (x-2)^2 + (y-3)^2 = 16 Therefore centre of circle is at (2, 3) and its radius is 4

Equation: x^2 +y^2 -4x -2y -4 = 0 Completing the squares: (x-2)^2 +(y-1)^2 -4-1-4 = 0 So: (x-2)^2 +(y-1)^2 = 9 Therefore the centre of the circle is at (2, 1) and its radius is 3

Centre = (0,0), the origin; radius = 11

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 )

No, It's a a quadratic equation because you have X squared.

(-4,3)

(-7,5)

It is the equation of a circle with radius of 6 and its center at the origin.

no, because xx=x squared, and x squared is not linear

Circle equation: x^2 +y^2 -8x +4y = 30 Tangent line equation: y = x+4 Centre of circle: (4, -2) Slope of radius: -1 Radius equation: y--2 = -1(x-4) => y = -x+2 Note that this proves that tangent of a circle is always at right angles to its radius

Equation of circle: x^2 +4x +y^2 -18y +59 = 0 Completing the squares: (x +2)^2 +(y -9)^2 = 26 which is radius squared Center of circle: (-2, 9) Area of circle: pi times 26 = 81.681 square units to three decimal places

area equals pi r squared therefor r squared equals area over pi. Now find square root of r squared and you have "R" (radius) = 2.821