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Practice
Circle, Circle Equations
Circle, Circle Equations: Problems with Solutions
Problem 1
Where is the center and what is the radius of the circle
$x^{2}+(y-3)^{2}=49$?
Graph the equation.
$C:(0,-3)\qquad r=49$
$C:(3,0)\qquad r=49$
$C:(-3,0)\qquad r=7$
$C:(0,3)\qquad r=7$
Solution:
Answer: Center at $(0,3)\qquad r=7$
The canonical form of the circle equation is
$(x-h)^{2}+(y-k)^{2}=r^{2}$, where center is at $C:(h, k)$ and radius $r$
We have $x^{2}+(y-3)^{2}=49,$
$C:(0,3)\qquad r=7$
Problem 2
Where is the center and what is the radius of the circle
$(x+2)^{2}+y^{2}=36$?
$C:(-2,0)\qquad r=6$
$C:(2,0)\qquad r=\sqrt{6}$
$C:(0,-2)\qquad r=\sqrt{6}$
$C:(0,2)\qquad r=6$
Solution:
The canonical form of the circle equation is
$(x-h)^{2}+(y-k)^{2}=r^{2}$, where the center is at $(h,k)$ and $r$ is its radius.
Our equation is
$(x+2)^{2}+y^{2}=36$
The center is at $(-2,0)$ and the radius is 6.
Problem 3
Complete the squares and determine the center and radius of the circle
$2x^{2}+2y^{2}+4x+16y+1=0$.
$C(-1,-4)$ and $r=\sqrt{\frac{33}{2}}$
$C(1,4)$ and $r=\sqrt{\frac{33}{2}}$
$C(4,1)$ and $r=17$
$C(4,-1)$ and $r=17$
Solution:
Answer: $C(-1,-4)$ and $r=\sqrt{\frac{33}{2}}$
$2x^{2}+2y^{2}+4x+16y+1=0\Longrightarrow x^{2}+y^{2}+2x+8y+\frac{1}{2}=0$ then
$\left( x+1\right)^{2}+\left( y+4\right)^{2}-1-16+\frac{1}{2}=0\Longrightarrow \left( x+1\right) ^{2}+\left( y+4\right) ^{2}=\frac{33}{2}$
We see that the center is at $(-1,-4)$ and the radius is $\sqrt{\frac{33}{2}}$
Problem 4
Which of the following is the equation of the circle with center at $(2,3)$ and radius $4$? Graph the equation.
$(x-2)^{2}+(y-3)^{2}=4$
$(x+2)^{2}+(y-3)^{2}=16$
$(x-2)^{2}+(y-3)^{2}=16$
$(x-2)^{2}+(y+3)^{2}=4$
Solution:
Answer: $(x-2)^{2}+(y-3)^{2}=16$
Canonical equation of a circle has the form
$(x-h)^{2}+(y-k)^{2}=r^{2}$, where the center is located at point $(h,k)$ and the radius is $r$
So the equation is $(x-2)^{2}+(y-3)^{2}=4^{2}\Longrightarrow (x-2)^{2}+(y-3)^{2}=16$
Problem 5
Which of the following is the equation of the circle with
center at $(-1, 4)$ and a radius of $4$? Graph the equation.
$(x+1)^{2}+(y-4)^{2}=4$
$(x-1)^{2}+(y-4)^{2}=4$
$(x+1)^{2}+(y+4)^{2}=4$
$(x-1)^{2}+(y+4)^{2}=4$
Solution:
Canonical equation of a circle has the form
$(x-h)^{2}+(y-k)^{2}=r^{2},$ where the center of the circle is located at point $(h,k)$ and the radius $r$
So $(x+1)^{2}+(y-4)^{2}=2^{2}\Longrightarrow (x+1)^{2}+(y-4)^{2}=4$
Problem 6
The center and radius of the circle
$x^{2}+y^{2}-4x+2y=0$ is:
$C(-2,-1)$, $r=\sqrt{5}$
$C(-2,1)$, $r=5$
$C(2,1)$, $r=5$
$C(2,-1)$, $r=\sqrt{5}$
Solution:
We need to convert the equation to sum of squares.
$x^{2}+y^{2}-4x+2y=0\Longrightarrow \left( x-2\right)^{2}+\left(y+1\right)^{2}-4-1=0$
Then $\left(x-2\right)^{2}+\left( y+1\right)^{2}=5$
We see that
the center is at $(2,-1)$ and the radius $r=\sqrt{5}$.
Problem 7
What is the equation of a circle with center at $(0,0)$ that passes through $(-1,-2)$?
$x^{2}-y^{2}=\sqrt{5}$
$x^{2}+y^{2}=\sqrt{5}$
$x^{2}-y^{2}=5$
$x^{2}+y^{2}=5$
Solution:
The canonical form of the equation of a circle is
$(x-h)^{2}+(y-k)^{2}=r^{2}$
The center is at point $(0,0)$ and $r$ is its radius.
So $x^{2}+y^{2}=r^{2}\Longrightarrow r=\sqrt{x^{2}+y^{2}}$
We know that the circle passes through $(-1,-2)$, therefore
$r=\sqrt{\left( -1\right)^{2}+\left(-2\right) ^{2}}=\sqrt{5}$
So the equation is $x^{2}+y^{2}=5$
Problem 8
What is the equation of a circle with center at $(4,-5)$ that passes through $(7,-3)$?
$(x-4)^{2}+(y+5)^{2}=13$
$(x+4)^{2}+(y+5)^{2}=13$
$(x-4)^{2}+(y-5)^{2}=13$
$(x-4)^{2}+(y-5)^{2}=13$
Solution:
The canonical form of the equation of a circle is
$(x-h)^{2}+(y-k)^{2}=r^{2},$ where the center is at point $(4,-5)$ and $r$ is its radius.
So $(x-4)^{2}+(y+5)^{2}=r^{2}\Longrightarrow r=\sqrt{(x-4)^{2}+(y+5)^{2}}$ Since the circle passes through $(7,-3)$ then
$r=\sqrt{\left( 3\right)^{2}+\left( 2\right)^{2}}=\sqrt{13}$
So the equation is $(x-4)^{2}+(y+5)^{2}=13$
Problem 9
A circle with center at $(5,6)$ touches the
x
-axis.
What is the equation of the circle?
$(x+5)^{2}+(y+6)^{2}=6$
$(x+5)^{2}+(y-6)^{2}=36$
$(x-5)^{2}+(y-6)^{2}=36$
$(x-5)^{2}+(y+6)^{2}=6$
Solution:
Answer: $(x-5)^{2}+(y-6)^{2}=36$
Since the circle touches the x-axis, $r=\left\vert y_{c}\right\vert$ where the center is $(x_{c},y_{c})$
So $y_{c}=6$ and the equation is $(x-5)^{2}+(y-6)^{2}=36$
Problem 10
A circle with center at $(-4,3)$ touches the
y
-axis.
What is the equation of the circle?
$(x-4)^{2}+(y-3)^{2}=4$
$(x+4)^{2}+(y+3)^{2}=16$
$(x+4)^{2}+(y-3)^{2}=16$
$(x-4)^{2}+(y+3)^{2}=4$
Solution:
Since the circle touches the y-axis, $r=\left\vert x_{c}\right\vert$ where the center is at $(x_{c},y_{c})$
So $x_{c}=-4$ and the equation is $(x+4)^{2}+(y-3)^{2}=16$
Problem 11
Find the equation of a circle with center $O(-2,3)$ and radius $r=4$. Which of the following points belongs to the circle?
$(2,3)$
$(-4,3)$
$(1,5)$
None of these
Solution:
Answer: $(2,3)$
$(x+2)^{2}+(y-3)^{2}=16$
Let we check which of the points satisfies this equation.
$A(2,3):(2+2)^{2}+(3-3)^{2}=16\Longrightarrow 16=16$ then $A(2,3)\in C$
$B(-4,3):(-4+2)^{2}+(3-3)^{2}=16\Longrightarrow 4\neq 16$ then $B(-4,3)\notin C$
$D(1,5):(1+2)^{2}+(5-3)^{2}=16\Longrightarrow 9+4=13\neq 16$ then $D(1,5)\notin C$
Problem 12
Which of the following lines is the tangent to the circle $x^{2}+y^{2}-4x-21=0$ at the point $T(5,4)$?
$4y-3x-31=0$
$4x+3y-31=0$
$4y+3x-31=0$
$4y+3x+31=0$
Solution:
Answer: $x^{2}+y^{2}+Dx+Ey+F=0$ where $D=-4\qquad E=0\qquad F=-21$
The equation of the tangent at the point $\left( x_{t},y_{t}\right)=(5,4)$ is
$y-y_{t}=\left( \frac{-2x_{t}-D}{2y_{t}+E}\right) (x-x_{t})$ So
$y-4=\left( \frac{-10+4}{8+0}\right) (x-5)\Longrightarrow y-4=\frac{-3}{4}(x-5)$
$4y+3x-31=0$
Problem 13
What is the equation of the circle with center $O(2,-1)$ that is tangent to the straight line $r:y=x+2$?
Graph the equation.
$(x+2)^{2}+(y+1)^{2}=\frac{25}{2}$
$(x-2)^{2}+(y-1)^{2}=\frac{25}{2}$
$(x-2)^{2}+(y+1)^{2}=\frac{25}{2}$
$(x+2)^{2}+(y-1)^{2}=\frac{25}{2}$
Solution:
Answer: $(x-2)^{2}+(y+1)^{2}=\frac{9}{5}$
We know that the radius $r$ is the distance between $O:(h,k)$ and the straight line $y-x-2=0$
So, $O:(2,-1)$ $r=d(O,L)=\frac{\left\vert -2-1-2\right\vert }{\sqrt{1^{2}+\left( -1\right) ^{2}}}=\frac{5}{\sqrt{2}}$ then the equation of the circle is
$(x-h)^{2}+(y-k)^{2}=r^{2}\Longrightarrow (x-2)^{2}+(y+1)^{2}=\frac{25}{2}$
Problem 14
What is the equation of the circle concentric with the circle $(x-2)^{2}+(y+1)^{2}=5$ that is tangent to the line $2x-y+2=0$?
$(x+2)^{2}+(y+1)^{2}=\frac{49}{5}$
$(x-2)^{2}+(y-1)^{2}=\frac{49}{5}$
$(x+2)^{2}+(y+1)^{2}=\frac{7}{5}$
$(x-2)^{2}+(y+1)^{2}=\frac{49}{5}$
Solution:
The concentric circle equation is $(x-2)^{2}+(y+1)^{2}=r^{2}$
We know that the radius $r$ is the distance between the center $O:(h,k)$ to line
So, $O:(2,-1)$ $r=d(C,L)=\frac{\left\vert 2\times 2+1+2\right\vert }{\sqrt{2^{2}+\left( -1\right) ^{2}}}=\frac{7}{\sqrt{5}}=\frac{7}{5}\sqrt{5}$
Then the circle equation is $(x-2)^{2}+(y+1)^{2}=\frac{49}{5}$
Problem 15
Which of the following straight lines is a tangent to the circle
$(x-2)^{2}+y^{2}=25$?
$L1:3x+4y-6=0$
$L2:3x+4y-31=0$
$L3:3x+4y-60=0$
$L4: 2x+y+25=0$
Solution:
Answer: L2: $3x+4y-31=0$
A straight line that is a tangent to a circle and the circle itself must have exactly one common point.
So for these lines we obtain $y=\frac{C-3x}{4}$, then substitute $y$ into
the equation of the circle so $(x-2)^{2}+\left( \frac{C-3x}{4}\right)^{2}=25$
Now we solve the equation,
$x^{2}-4x+4+\frac{C^{2}-6Cx+9x^{2}}{16}=25\Longrightarrow 16x^{2}-64x+64+C^{2}-6Cx+9x^{2}=400$
Then $25x^{2}-(64+6C)x+\left( 64+C^{2}-400\right) =0$
$x=\frac{(64+6C)\pm \sqrt{(64+6C)^{2}-100\left( 64+C^{2}-400\right) }}{50}$
Since $x$ has only one solution (the point of tangency),
$(64+6C)^{2}-100\left( 64+C^{2}-400\right) =0$ We have three options for $C=6,31,60$
The unique value $C$ which satisfies the previous expression is $C=31$ since
$(64+6\cdot 31)^{2}-100\left( 64+31^{2}-400\right) =250^{2}-62500=0$
So L2: $3x+4y-31=0$ is the straight line tangent to the circle.
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