Question

The diameter of Circle $$Q$$ terminates on the circumference of the circle at $$(5,-1)$$ and $$(-5,1)$$. Write the equation of the circle in standard form. Show all of your work.

1. thienhuong
The equation of the given circle in the standard form is given as x² + y² = 26.
In the question, we are given that the diameter of Circle $$Q$$ terminates on the circumference of the circle at $$(5,-1)$$ and $$(-5,1)$$.
We are asked to write the equation of the circle in standard form.
The equation of the circle in standard form is given as (x – h)² + (y – k)² = r, where (h, k) is the center of the circle, and r is the radius of the circle.
The center of the circle can be found using the midpoint formula, as the center is the midpoint of the diameter.
Thus, the center is at ( (5 + (-5))/2, (-1 + 1)/2 ) = (0,0).
The length of the diameter can be calculated using the distance formula.
Thus, the length of the diameter of the circle is √((-5 – 5)² + (1 – (-1))²) = √( (-10)² + (2)² ) = √(100 + 4) = √104 = 2√26.
The radius is half the length of the diameter.
Thus, the radius, r = √26.
Now, we have the center (h, k) = (0,0), and the radius, r = √26.
Thus, the equation of the circle in the standard form, (x – h)² + (y – k)² = r ² can be shown as:
(x – 0)² + (y – 0)² = (√26)²,
or, x² + y² = 26.
Thus, the equation of the given circle in the standard form is given as x² + y² = 26.