9.5 Write equations of circles in standard form using properties EXA
Write the equation in standard form for the circle passing through (-2, 6) centered at the
origin.
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Answer :

To write the equation of a circle in standard form, we use the formula: (x - h)^2 + (y - k)^2 = r^2 where (h, k) is the center of the circle and r is the radius. Since the circle is centered at the origin, the values of h and k are both 0. Thus, our equation will simplify to: (x - 0)^2 + (y - 0)^2 = r^2 x^2 + y^2 = r^2 To determine the radius r, we need to use the distance formula to find the distance between the center of the circle at the origin (0, 0) and the point (-2, 6) which lies on the circle. The distance formula is: r = sqrt((x2 - x1)^2 + (y2 - y1)^2) Here, (x1, y1) is the center of the circle, and (x2, y2) is the point the circle passes through. By plugging in the values: r = sqrt((-2 - 0)^2 + (6 - 0)^2) r = sqrt((-2)^2 + (6)^2) r = sqrt(4 + 36) r = sqrt(40) r = sqrt(4 * 10) r = 2 * sqrt(10) (We can pull out a square factor of 4 out of the square root as 2.) Now, we have found the radius, which is 2 * sqrt(10). The standard form of the circle's equation with this radius is: x^2 + y^2 = (2 * sqrt(10))^2 When we square the radius, we square both the 2 and the sqrt(10): x^2 + y^2 = (2)^2 * (sqrt(10))^2 x^2 + y^2 = 4 * 10 x^2 + y^2 = 40 The standard form equation for the circle passing through (-2, 6) and centered at the origin is therefore: x^2 + y^2 = 40