The distance formula comes from the Pythagorean theorem: the horizontal leg is \(|x_2-x_1|\), the vertical leg is \(|y_2-y_1|\), and the hypotenuse is the distance.
⭕
Circle — Standard Form
MEMORIZE
Standard form
\((x - h)^2 + (y - k)^2 = r^2\)
→Center: \((h,\ k)\)
→Radius: \(r\)
→If center is at origin: \(x^2 + y^2 = r^2\)
→Watch the sign! \((x-3)^2 + (y+2)^2 = 16\) has center \((3,\,-2)\), \(r = 4\)
What is the midpoint of the segment joining \((2,\ 6)\) and \((8,\ -2)\)?
📘 Solution
\(M = \left(\dfrac{2+8}{2},\ \dfrac{6+(-2)}{2}\right) = \left(5,\ 2\right)\) Answer: B
4
Standard Form
EASY
What is the center and radius of the circle \((x-3)^2+(y+2)^2=16\)?
📘 Solution
Standard form \((x-h)^2+(y-k)^2=r^2\) gives \(h=3\), \(k=-2\) (note the sign in \(y+2\) means \(k=-2\)), and \(r=\sqrt{16}=4\). Answer: C
5
Writing Equations
EASY
Which equation represents the circle centered at the origin with radius 7?
📘 Solution
Circle at origin: \(x^2+y^2=r^2=7^2=49\).
Common error: writing \(r\) instead of \(r^2\). Always square the radius. Answer: B
6
Point & Circle
EASY
Where does the point \((5,\ 0)\) lie in relation to the circle \(x^2+y^2=25\)?
📘 Solution
Substitute: \(5^2+0^2 = 25\). Since the result equals 25 (= \(r^2\)), the point lies on the circle. Answer: A
7
General Form
MEDIUM
Find the center and radius of \(x^2+y^2-4x+6y-3=0\).
📘 Solution
Complete the square:
\((x^2-4x+4)+(y^2+6y+9) = 3+4+9 = 16\)
\((x-2)^2+(y+3)^2=16\)
Center \((2,\ -3)\), \(r = \sqrt{16} = 4\). Answer: B
8
Circle from Center & Point
MEDIUM
A circle has center \((1,\ 2)\) and passes through \((4,\ 6)\). What is its equation?
📘 Solution
\(r = \sqrt{(4-1)^2+(6-2)^2} = \sqrt{9+16} = \sqrt{25} = 5\)
\(r^2 = 25\), so the equation is \((x-1)^2+(y-2)^2=25\). Answer: B
9
Point-to-Line Distance
MEDIUM
What is the distance from the point \((2,\ 3)\) to the line \(3x+4y-1=0\)?
📘 Solution
\(d = \dfrac{|3(2)+4(3)-1|}{\sqrt{3^2+4^2}} = \dfrac{|6+12-1|}{\sqrt{25}} = \dfrac{17}{5}\) Answer: B
10
Line–Circle Intersection
MEDIUM
How many points does the line \(y=x+1\) share with the circle \(x^2+y^2=25\)?
📘 Solution
Substitute \(y=x+1\) into the circle equation:
\(x^2+(x+1)^2=25 \Rightarrow 2x^2+2x+1=25 \Rightarrow 2x^2+2x-24=0\)
\(x^2+x-12=0 \Rightarrow (x+4)(x-3)=0\)
\(x=-4\) or \(x=3\) → two intersection points: \((-4,\,-3)\) and \((3,\,4)\). Answer: C
11
Standard → General Form
MEDIUM
Which of the following is the general form of \((x+2)^2+(y-5)^2=9\)?
📘 Solution
Expand: \((x^2+4x+4)+(y^2-10y+25)=9\)
\(x^2+y^2+4x-10y+29=9\)
\(x^2+y^2+4x-10y+20=0\) Answer: B
12
Inside / Outside
MEDIUM
Which of the following points lies outside the circle \((x-1)^2+(y-1)^2=10\)?
📘 Solution
Substitute each point into \((x-1)^2+(y-1)^2\) and compare to 10:
A: \((4-1)^2+(4-1)^2=9+9=18>10\) → outside
B: \((0-1)^2+(0-1)^2=1+1=2<10\) → inside
C: \((1-1)^2+(4-1)^2=0+9=9<10\) → inside
D: \((-2-1)^2+(1-1)^2=9+0=9<10\) → inside Answer: A
13
General Form — Radius
MEDIUM
What is the radius of the circle \(x^2+y^2-6x+8y=0\)?
📘 Solution
Complete the square:
\((x^2-6x+9)+(y^2+8y+16)=0+9+16=25\)
\((x-3)^2+(y+4)^2=25\)
\(r = \sqrt{25} = 5\) Answer: C
14
Distance Between Parallel Lines
MEDIUM
What is the distance between the parallel lines \(3x+4y=5\) and \(3x+4y=15\)?
📘 Solution
Distance between parallel lines \(ax+by=c_1\) and \(ax+by=c_2\):
\(d = \dfrac{|c_1-c_2|}{\sqrt{a^2+b^2}} = \dfrac{|5-15|}{\sqrt{9+16}} = \dfrac{10}{5} = 2\) Answer: A
15
Tangent to Axis
MEDIUM
A circle has center \((3,\ 4)\) and is tangent to the \(x\)-axis. What is its equation?
📘 Solution
When a circle is tangent to the \(x\)-axis, the radius equals the absolute value of the \(y\)-coordinate of the center.
\(r = |4| = 4\), so \(r^2 = 16\).
Equation: \((x-3)^2+(y-4)^2=16\). Answer: B
16
Point on Circle
MEDIUM
If the point \((2,\ k)\) lies on the circle \(x^2+y^2=29\), what are the possible values of \(k\)?
📘 Solution
Substitute \(x=2\): \(4 + k^2 = 29 \Rightarrow k^2 = 25 \Rightarrow k = \pm 5\). Answer: B
17
Two Circles — Position
HARD
Consider the circles \((x-2)^2+y^2=9\) and \((x-6)^2+y^2=1\). How do they relate to each other?
📘 Solution
Circle 1: center \((2,0)\), \(r_1=3\). Circle 2: center \((6,0)\), \(r_2=1\).
Distance between centers: \(d = \sqrt{(6-2)^2+0^2} = 4\)
Since \(d = 4 = r_1+r_2 = 3+1\), the circles are externally tangent. Answer: B
18
Area of Circle
HARD
What is the area of the circle \(x^2+y^2-2x-4y-4=0\)?
📘 Solution
Complete the square:
\((x^2-2x+1)+(y^2-4y+4)=4+1+4=9\)
\((x-1)^2+(y-2)^2=9\)
\(r^2=9\), so area \(= \pi r^2 = 9\pi\). Answer: C
19
Minimum Distance
HARD
What is the minimum distance from the origin to a point on the circle \((x+1)^2+(y-2)^2=25\)?
📘 Solution
Center \((-1,2)\), \(r=5\).
Distance from origin to center: \(d = \sqrt{(-1)^2+2^2} = \sqrt{5}\)
Since \(\sqrt{5} \approx 2.24 < 5 = r\), the origin is inside the circle.
Minimum distance from origin to the circle \(= r - d = 5 - \sqrt{5}\). Answer: A
20
Diameter → Circle
HARD
A circle has diameter endpoints \(A(2,\ 0)\) and \(B(6,\ 4)\). Which equation represents this circle?
📘 Solution
Center = midpoint of diameter: \(\left(\dfrac{2+6}{2},\ \dfrac{0+4}{2}\right) = (4,\ 2)\)
Radius = half the diameter: \(r = \dfrac{1}{2}\sqrt{(6-2)^2+(4-0)^2} = \dfrac{1}{2}\sqrt{32} = \dfrac{4\sqrt{2}}{2} = 2\sqrt{2}\)
\(r^2 = 8\), so equation: \((x-4)^2+(y-2)^2=8\). Answer: B