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Zorluk: OrtaEquations of Circles in the Coordinate Plane

In the xyxy-plane, the graph of the equation x2+y2−8x+6y−11=0x^2 + y^2 - 8x + 6y - 11 = 0 is a circle. If the center of the circle is (h,k)(h, k) and the radius is rr, what is the value of h+k+rh + k + r?

Cevap: 7

Cevap

7
To find h+k+rh + k + r, we convert the general equation x2+y2−8x+6y−11=0x^2 + y^2 - 8x + 6y - 11 = 0 to standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 by completing the square. Grouping terms gives (x2−8x)+(y2+6y)=11(x^2 - 8x) + (y^2 + 6y) = 11. Adding 16 and 9 to both sides to complete the square yields (x2−8x+16)+(y2+6y+9)=11+16+9(x^2 - 8x + 16) + (y^2 + 6y + 9) = 11 + 16 + 9, which simplifies to (x−4)2+(y+3)2=36(x - 4)^2 + (y + 3)^2 = 36. Comparing this to the standard form gives center (h,k)=(4,−3)(h, k) = (4, -3) and radius r=36=6r = \sqrt{36} = 6. The sum of these values is 4+(−3)+6=74 + (-3) + 6 = 7.

Adım Adım Çözüm

1
Group the variable terms and move the constant term to the right side of the equation.
(x2−8x)+(y2+6y)=11(x^2 - 8x) + (y^2 + 6y) = 11
To arrange the equation for completing the square.
2
Complete the square for both variables by adding the square of half the coefficient of the linear terms to both sides.
(x2−8x+16)+(y2+6y+9)=11+16+9(x^2 - 8x + 16) + (y^2 + 6y + 9) = 11 + 16 + 9
This creates factorable perfect square trinomials on the left side while maintaining equality.
3
Factor the perfect square trinomials and add the constants on the right side.
(x−4)2+(y+3)2=36(x - 4)^2 + (y + 3)^2 = 36
To write the equation in the standard form of a circle: (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2.
4
Determine the coordinates of the center (h,k)(h, k) and the radius rr from the standard form.
Center coordinates h=4h = 4, k=−3k = -3, and radius r=6r = 6
Comparing (x−4)2+(y+3)2=36(x - 4)^2 + (y + 3)^2 = 36 to standard form reveals h=4h = 4, k=−3k = -3, and r=36=6r = \sqrt{36} = 6.
5
Sum the values of hh, kk, and rr together.
4 + (-3) + 6 = 7
To find the final requested value.

Anahtar Kavram

Converting the general equation of a circle into standard form by completing the square to identify its center and radius.
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Konular

Equations of Circles in the Coordinate Plane

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