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Zorluk: OrtaEquations and Graphs of Circles

In the standard (x,y)(x, y) coordinate plane, a circle is defined by the equation x2+y2−6x+8y−11=0x^2 + y^2 - 6x + 8y - 11 = 0. What is the distance from the center of this circle to the origin (0,0)(0, 0)?

Cevap: 5

Cevap

The distance from the center of the circle to the origin is 5.
By completing the square on x2−6xx^2 - 6x and y2+8yy^2 + 8y, we rewrite x2+y2−6x+8y−11=0x^2 + y^2 - 6x + 8y - 11 = 0 as (x−3)2+(y+4)2=36(x - 3)^2 + (y + 4)^2 = 36. The center of the circle is (3,−4)(3, -4). The distance from (3,−4)(3, -4) to (0,0)(0, 0) is 32+(−4)2=25=5\sqrt{3^2 + (-4)^2} = \sqrt{25} = 5.

Adım Adım Çözüm

1
Rewrite the general equation of the circle in standard form by completing the square.
(x−3)2+(y+4)2=36(x - 3)^2 + (y + 4)^2 = 36
Grouping xx and yy terms and adding (b/2)2(b/2)^2 to both sides isolates the center coordinates (h,k)(h, k).
2
Identify the center of the circle from standard form.
Center is (3,−4)(3, -4)
The standard equation of a circle is (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the center.
3
Compute the distance between the center (3,−4)(3, -4) and the origin (0,0)(0, 0).
d=32+(−4)2=25=5d = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5
Applying the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} yields the distance to the origin.

Anahtar Kavram

Completing the square to find the center of a circle and applying the distance formula
Tahmini Süre:1m 15s
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Konular

Equations and Graphs of Circles

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