Examkin

Soru

Zorluk: OrtaEquations of Circles in the Coordinate Plane

In the xyxy-plane, the graph of the equation x2+y2−10x−24y+69=0x^2 + y^2 - 10x - 24y + 69 = 0 is a circle. What is the distance between the center of this circle and the origin?

Cevap: 13

Cevap

The distance between the center of the circle and the origin is 13.
Completing the square on the given equation yields standard form (x−5)2+(y−12)2=100(x - 5)^2 + (y - 12)^2 = 100, identifying the center of the circle as (5,12)(5, 12). The distance from (5,12)(5, 12) to (0,0)(0,0) is calculated using the distance formula: 52+122=169=13\sqrt{5^2 + 12^2} = \sqrt{169} = 13.

Adım Adım Çözüm

1
Group the xx and yy terms and complete the square for each variable.
(x−5)2+(y−12)2=100(x - 5)^2 + (y - 12)^2 = 100
By rewriting x2−10xx^2 - 10x as (x−5)2−25(x - 5)^2 - 25 and y2−24yy^2 - 24y as (y−12)2−144(y - 12)^2 - 144, the equation becomes (x−5)2−25+(y−12)2−144+69=0(x - 5)^2 - 25 + (y - 12)^2 - 144 + 69 = 0. Combining the constant terms gives (x−5)2+(y−12)2−100=0(x - 5)^2 + (y - 12)^2 - 100 = 0, which simplifies to standard form.
2
Identify the center of the circle from the standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2.
The center is (5,12)(5, 12).
Comparing (x−5)2+(y−12)2=100(x - 5)^2 + (y - 12)^2 = 100 to the standard form shows that h=5h = 5 and k=12k = 12.
3
Use the distance formula to calculate the distance between the center (5,12)(5, 12) and the origin (0,0)(0, 0).
13
The distance formula is d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. Substituting the points gives d=(5−0)2+(12−0)2=25+144=169=13d = \sqrt{(5 - 0)^2 + (12 - 0)^2} = \sqrt{25 + 144} = \sqrt{169} = 13.

Anahtar Kavram

Equations of Circles in the Coordinate Plane
Bu soruyu puanla

Konular

Equations of Circles in the Coordinate Plane

Bu konudan daha fazla soru