This question asks for the area of a circle with center (1, -5) passing through (1, 2). First, find the radius as the distance between these points: $√[(1-1)^2$ + (2 - $(-5))^2$] = √[0 + $7^2$] = 7. The area formula is $πr^2$ = π × 49 = 49π square units, using the correct radius. Note the vertical distance since x-coordinates match. A common error is squaring the radius incorrectly or using diameter. In exams, verify by substituting back to ensure the point lies on the circle.

A chord is 10 units long with perpendicular distance 12 units from the center. When a perpendicular from the center meets a chord, it bisects the chord, creating a right triangle with legs 5 (half the chord) and 12 (perpendicular distance), and hypotenuse r (radius). Using the Pythagorean theorem: r² = 5² + 12² = 25 + 144 = 169, so r = 13 units. The key insight is that the perpendicular from the center always bisects the chord. This problem tests understanding of the perpendicular bisector property of chords.

The equation is in standard form (x-h)² + (y-k)² = r², where (h,k) is the center and r is the radius. From (x-3)² + (y+5)² = 64, we identify: center = (3, -5) and r² = 64, so r = 8. Note that y+5 = y-(-5), so the y-coordinate of the center is -5, not 5. A common error is misreading the signs in the equation. When identifying the center from standard form, remember that (x-h) means the x-coordinate is h, and (y-k) means the y-coordinate is k.

This question asks to identify the center and radius from the standard form equation of a circle. The standard form is (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius. From (x - 3)² + (y + 2)² = 49, we can see the center is (3, -2) since we have (x - 3) and (y - (-2)). The radius is √49 = 7, not 49 itself. Common mistakes include misidentifying the signs of the center coordinates or confusing r² with r.

This problem asks for the arc length intercepted by a 150° central angle in a circle with radius 8 cm. The formula for arc length is s = (θ/360°) × 2πr when θ is in degrees. Substituting: s = (150°/360°) × 2π(8) = (5/12) × 16π = 80π/12 = 20π/3 cm. A common error is forgetting to convert the angle fraction or miscalculating the simplification. Always check that your angle is in the correct units for your formula.

This question involves a fundamental property of tangent lines to circles. A tangent line to a circle is always perpendicular to the radius drawn to the point of tangency. Since OP is the radius to point P where the line is tangent, the angle between OP and the tangent line must be 90°. The given information that the distance from O to the tangent line equals OP = 13 confirms this is the perpendicular distance. Remember this key property for any tangent line problem.

This question asks for the circumference of a circle with diameter 18 ft. The circumference formula is C = πd when using diameter, or C = 2πr when using radius. Since diameter = 18 ft, the circumference is C = π(18) = 18π ft. A common mistake is confusing diameter with radius and using C = 2π(18) = 36π, which would double the correct answer. Remember that diameter is already twice the radius.

This problem asks for the equation of a circle with center (2, -1) that passes through (8, 3). The standard form is (x - h)² + (y - k)² = r², where we need to find r². The radius is the distance from center to the given point: r² = (8 - 2)² + (3 - (-1))² = 6² + 4² = 36 + 16 = 52. Therefore, the equation is (x - 2)² + (y + 1)² = 52. Be careful with signs when substituting the center coordinates into the standard form.

This question asks for the area of a pizza slice with a 45° central angle from a pizza with radius 7 inches. The formula for sector area is $A = \frac{\theta}{360^\circ} \times \pi r^2$. Substituting: $A = \frac{45^\circ}{360^\circ} \times \pi(7)^2 = \frac{1}{8} \times 49\pi = \frac{49\pi}{8}$ square inches. Common mistakes include using diameter instead of radius or incorrectly simplifying the fraction 45/360 to something other than 1/8. When dealing with pizza or pie problems, visualize the fraction of the whole circle.

This problem asks for the arc length when a runner completes $\frac{3}{8}$ of a circular track. The formula for arc length is $s = r \theta$, where $\theta$ is in radians, or $s = (\text{fraction of circle}) \times \text{circumference}$. The circumference is $C = 2\pi r = 2\pi(14) = 28\pi$ meters. For $\frac{3}{8}$ of the lap, the distance is $(\frac{3}{8}) \times 28\pi = \frac{84\pi}{8} = \frac{21\pi}{2}$ meters. A common error is forgetting to multiply by $2\pi$ when finding circumference, which would give half the correct answer.