Write the equation of the circle in standard form. x^2+y^2-6x+8y+9=0.

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The expression x^2 + y^2 - 6x + 8y + 9 = 0 is the equation of a circle.

The standard form of a circle is (x - a)^2 + (y - b)^2 = r^2, where r is the radius of the circle and (a,b) is the center.

x^2 + y^2 - 6x + 8y + 9 = 0

=> x^2 - 6x + y^2 + 8y + 9 = 0

=> x^2 - 6x + 9 + y^2 + 8y + 16 = 9 + 16 - 9

=> (x - 3)^2 + (y + 4)^2 = 4^2

The standard form of the equation of the circle is (x - 3)^2 + (y + 4)^2 = 4^2

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll recall the standard form of the circle equation:

(x - h)^2 + (y - k)^2 = r^2

h and k represent the coordinates of the center of the circle and r is the value of the radius of the circle.

To reach to this form, we'll have to complete the squares:

(x^2 - 6x + ...) + (y^2 + 8y + ...) = -9

We'll consider the formula:

(a+b)^2 = a^2 + 2ab + b^2

2xb = -6x

b = -3 => b^2 = 9

x^2 - 6x + ... = x^2 - 6x + 9

y^2 + 8y + ... = y^2 + 8y + 16

(x^2 - 6x + 9) + (y^2 + 8y + 16)  -9 -16 = -9

We'll move the numbers to the right side:

(x^2 - 6x + 9) + (y^2 + 8y + 16) = 9 + 16 - 9

(x^2 - 6x + 9) + (y^2 + 8y + 16) = 16

(x-3)^2 + (y + 4)^2 = 4^2

The standard form of the equation of the circle is: (x-3)^2 + (y + 4)^2 = 4^2

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