This collection of 20 circle geometry problems covers the distance formula, the equation of a circle, finding the center and radius, tangent lines, and points of intersection. Every problem includes a full solution — perfect for high school and introductory precalculus students.
Find the distance between the points \( A(3,4) \) and \( B(5,8) \).
Use the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):
\[ d = \sqrt{(5 - 3)^2 + (8 - 4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]Find all values of \( x \) such that the distance between \( (x,4) \) and \( (-5,3) \) equals \( 5 \).
Note: older answer keys listing \( x = -7 \) or \( x = -3 \) are incorrect.
Write the equation of the circle with center \( (0,0) \) and radius \( 6 \).
Standard form \( (x-h)^2 + (y-k)^2 = r^2 \) with \( h=0, k=0, r=6 \):
\[ x^2 + y^2 = 36 \]Find the center and radius of \( (x - 2)^2 + (y + 5)^2 = 13 \).
Comparing with \( (x-h)^2 + (y-k)^2 = r^2 \):
Center \( (2,-5) \), radius \( \sqrt{13} \).
Find the center and radius of \( (5 - x)^2 + (y - 1)^2 = 4 \).
Since \( (5-x)^2 = (x-5)^2 \):
\[ (x - 5)^2 + (y - 1)^2 = 4 \]Center \( (5,1) \), radius \( 2 \).
Find the center and radius of \( (-4 - x)^2 + (-y + 11)^2 = 9 \).
Factoring out \(-1\) inside each square:
\[ (x + 4)^2 + (y - 11)^2 = 9 \]Center \( (-4,11) \), radius \( 3 \).
Find the center and radius of \( x^2 + y^2 + 6x - 10y = 9 \).
Center \( (-3,5) \), radius \( \sqrt{43} \).
Find the center and radius of \( x^2 + y^2 + 4y = 0 \).
Center \( (0,-2) \), radius \( 2 \).
Find the center and radius of \( -x^2 - y^2 + 8x = 0 \).
Multiply by \(-1\): \( x^2 + y^2 - 8x = 0 \).
\[ (x-4)^2 + y^2 = 16 \]Center \( (4,0) \), radius \( 4 \).
Find the equation of the circle with diameter endpoints \( (-6,1) \) and \( (2,-5) \).
Center = midpoint \( (-2,-2) \).
\[ r = \sqrt{(2+2)^2 + (-5+2)^2} = \sqrt{16+9} = 5 \]Equation: \( (x+2)^2 + (y+2)^2 = 25 \).
Find the intersection points of \( x^2 + y^2 = 4 \) and \( (x-2)^2 + (y-2)^2 = 4 \).
Expanding the second circle and substituting \( x^2+y^2=4 \):
\[ 8 - 4x - 4y = 0 \implies y = 2 - x \]Substituting back: \( x^2 + (2-x)^2 = 4 \implies 2x(x-2)=0 \).
Intersection points: \( (2,0) \) and \( (0,2) \).
Find the equation of the circle with center \( (-3,5) \) passing through \( (5,-1) \).
Equation: \( (x+3)^2 + (y-5)^2 = 100 \).
Find the equation of the circle through \( (0,6) \), \( (0,0) \), and \( (8,0) \).
Right triangle at the origin → hypotenuse is the diameter from \( (0,6) \) to \( (8,0) \).
Center \( (4,3) \), \( r^2 = 4^2 + 3^2 = 25 \).
Equation: \( (x-4)^2 + (y-3)^2 = 25 \).
Find the intersection of \( (x-2)^2 + y^2 = 8 \) and \( y = x - 2 \).
Points: \( (0,-2) \) and \( (4,2) \).
For which \( K \) is \( y = Kx \) tangent to \( (x-7)^2 + (y-6)^2 = 9 \)?
Set discriminant to zero:
\[ 40K^2 - 84K + 27 = 0 \] \[ K = \frac{21 \pm 3\sqrt{19}}{20} \]For which \( K \) is \( x = 6 \) tangent to \( x^2 + y^2 = K \)?
Tangency requires a single solution for \( y \), so \( y^2 = 0 \):
\[ K = 36 \]Is \( (3,5) \) inside, outside, or on \( x^2 + y^2 = 9 \)?
Outside.
For which \( K \) is \( (1,K) \) inside \( x^2 + y^2 = 4 \)?
\( -\sqrt{3} < K < \sqrt{3} \).
For which \( K \) is \( (K,-2) \) outside \( x^2 + y^2 = 9 \)?
\( K \in (-\infty,-\sqrt{5}) \cup (\sqrt{5},\infty) \).
For which \( K \) is \( (K,2K) \) on \( x^2 + y^2 = 5 \)?
\( K = \pm 1 \).
\( (x-h)^2 + (y-k)^2 = r^2 \), where \( (h,k) \) is the center and \( r \) is the radius.
Complete the square for both \( x \) and \( y \) to rewrite the equation in standard form.
Substitute the line into the circle's equation. If the resulting quadratic has a discriminant of zero, the line is tangent.
Compare \( x^2 + y^2 \) with \( r^2 \): less means inside, equal means on, greater means outside.