12 multiple-choice questions, progressively harder.
Which condition describes every point of a circle with center (h,k)(h,k)(h,k) and positive radius rrr?
Solution
Correct answer: B
A point belongs to the circle exactly when its distance from the center is the positive radius. Squaring that distance gives
The strict inequality describes the interior. A zero right side describes only the center, and changing the signs and using r instead of its square gives a different condition.
Which equation describes a circle with center (3,−5)(3,-5)(3,−5) and radius 6\sqrt66?
Correct answer: D
Subtract each center coordinate inside its square. The square of the radius is 6, so
The sign-reversed equation has a different center. Neither 36 nor the square root of 6 is the squared radius.
Which point lies on (x+2)2+(y−4)2=10(x+2)^2+(y-4)^2=10(x+2)2+(y−4)2=10?
Correct answer: A
Substitute the coordinates of each point into the left side. For the point (1,5),
The other points give 0, 4, and 13 respectively. Only 10 equals the squared radius.
What is the rightmost point of (x+6)2+(y+4)2=25(x+6)^2+(y+4)^2=25(x+6)2+(y+4)2=25?
Correct answer: C
The center is (-6,-4) and the radius is 5. Move 5 units right from the center.
The first coordinate is negative 1 and the height stays negative 4. The other choices move left, move up, or mistake 25 for the radius.
What is the real graph of (x+7)2+(y−4)2=0(x+7)^2+(y-4)^2=0(x+7)2+(y−4)2=0?
Both real squares are nonnegative, so their sum is zero only if both vanish.
This gives the single point (-7,4). Its distance from itself is zero, so it is not a positive-radius circle.
Why does (x−1)2+(y−6)2=−4(x-1)^2+(y-6)^2=-4(x−1)2+(y−6)2=−4 have no real graph?
For real coordinates each square is nonnegative. Thus
It cannot equal negative 4. There are no real points, rather than a circle with a negative radius or a pair of points.
Where is (2,1)(2,1)(2,1) relative to (x+1)2+(y−2)2=13(x+1)^2+(y-2)^2=13(x+1)2+(y−2)2=13?
The center is (-1,2). The squared distance to (2,1) is
Since 10 is less than 13, the point is inside. It differs from the center, whose squared center distance is zero.
Find every common point of the circle and line.
Substitute the fixed height into the circle.
The first coordinate can be negative 4 or 4. Each pair has height negative 3 and squared coordinate sum 25. The swapped-coordinate choices do not satisfy the line.
How many common points do the circle and line have?
Substitute the fixed first coordinate.
The only height is 3, giving the single point (-4,3). The circle has positive radius 2, so this line is tangent.
Substitute the line height.
No real first coordinate works, so the line misses the circle.
A circle has center (2,1)(2,1)(2,1) and passes through T=(2,−3)T=(2,-3)T=(2,−3). Which line is tangent at T?
The contact radius is vertical because the center and T share first coordinate 2. The perpendicular through T is horizontal.
The first two alternatives pass through the center. The line x equals 6 does not pass through T.
T is on a positive-radius circle with center C, and T is also on a line. Which additional fact proves the line tangent at T?
The tangent-radius converse applies at a point of the circle. Its needed direction condition is
A line through the center is a secant. Parallelism does not supply the required right angle, and a defined slope alone says nothing about its angle to CT.
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