Circles in the Coordinate Plane Practice Quiz

The standard form of a circle's equation is (x - h)² + (y - k)² = r² where (h, k) represents the center and r is the radius. This form clearly shows the geometric features of the circle.

In the equation, h and k designate the x and y coordinates of the circle's center. This directly identifies the location of the circle on the coordinate plane.

The radius is found by taking the square root of 16, which results in 4. This follows directly from the standard form where r² is on the right side of the equation.

The expression (x + 1)² can be rewritten as (x - (-1))², showing that the x-coordinate of the center is -1. Similarly, (y - 5)² indicates that the y-coordinate is 5, so the center is (-1, 5).

A point lies on the circle if, when substituted into the equation, the equality holds true. This differentiates points on the circle from those inside or outside it.

The standard form (x - h)² + (y - k)² = r² explicitly shows the center as (h, k) and the radius as the square root of the right-hand side. Completing the square is necessary only when converting from the general form.

Completing the square for both x and y transforms the equation into (x + 3)² + (y - 2)² = 4, which reveals the center as (-3, 2). This method is standard for converting the general form into the standard form.

Since the equation shows that r² = 25, the radius is the positive square root of 25, which is 5. This direct calculation confirms the circle's size.

Substituting (5, 6) into the equation yields (5 + 2)² + (6 - 6)² = 7² + 0² = 49, which satisfies the circle's equation. The other options do not meet this condition.

Expanding (x - h)² gives x² - 2hx + h², so the linear term in x is -2hx. This term connects the standard form to the general form of the circle's equation.

Converting from the general to the standard form requires completing the square for both x and y terms. This method restructures the equation to clearly display the center and radius.

Using the distance formula, the distance is calculated as √[(4 - (-2))² + (5 - 1)²] = √(6² + 4²) = √(36 + 16) = √52, which simplifies to 2√13. This confirms the distance between the two points.

Since the area of a circle is given by πr², doubling the radius results in an area of π(2r)² = 4πr². Thus, the area becomes four times as large.

A circle is tangent to the x-axis if the absolute value of the y-coordinate of its center equals the radius. For (x - 1)² + (y - 3)² = 9, the center is (1, 3) and the radius is 3, which satisfies this condition.

The circle's center is at (2, 2) and moving it to the origin requires subtracting 2 from both coordinates, which means a translation by (-2, -2). Other transformations do not achieve the required shift.

Completing the square for x and y transforms the equation into (x + 4)² + (y - 5)² = 36, revealing the center as (-4, 5) and the radius as √36, which is 6. This process is standard for rewriting a circle's equation into its standard form.

A line is tangent to a circle if the perpendicular distance from the circle's center to the line equals the radius. The distance formula |mh - k + b|/√(m²+1) set equal to r provides the necessary condition.

The perpendicular distance from the center (a, 3) to the line y = x is |a - 3|/√2. Setting this equal to the radius 4 yields |a - 3| = 4√2, so one possible solution is a = 3 + 4√2. Although another solution exists, only one option is marked correct.

For externally tangent circles, the sum of the radii equals the distance between the centers. The distance between (-2, 4) and (1, -1) is √34, so the radius of the circle with center (-2, 4) is √34 - 5, leading to the equation (x + 2)² + (y - 4)² = (√34 - 5)².

By completing the square, the general form converts to the standard form with the radius squared given by (D/2)² + (E/2)² - F. This value must be greater than zero for the circle to have a positive radius.