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.
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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.Learn more about the equation of the circle athttps://brainly.com/question/28192971#SPJ4