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This is the second of our many quizzes on Circle equations. Click on this to access more such competitive quizzes on Geometry.

Circle equations Quiz 2

mathematics

 If the circle has a radius of $24$ units and a diameter coincides that with the $x-$ axis, calculate the possible equations of a circle such that the end of the circle’s diameter touches the origin. 

 Calculate the length of the diameter using the equation $D=2r$. Identify the equations by considering two points of the end of a diameter and use them to calculate the center of the respective circle. Plug in the values of the radius and the center of the circles into the equation of a circle. 

Radius: $24$ units Origin: $\left( 0,0 \right)$ $D=2r \\ \Rightarrow D=2\times 24 \\ \Rightarrow D=48 $ $\text{1st}$ equation: The circle is present to the right of the $y-$ axis $\text{1st}$ end of the diameter: $\left( 0,0 \right)$ $2\text{nd}$ end of the diameter: $\left( 48,0 \right)$ Midpoint $=\left( \dfrac{0+48}{2},\dfrac{0+0}{2} \right)=\left( 24,0 \right)$ Therefore, the center of the circle has the coordinates: $\left( 24,0 \right)$ ${{\left( x-24 \right)}^{2}}+{{y}^{2}}=576$ $2\text{nd}$ equation: The circle is present to the left of the $y-$ axis $\text{1st}$ end of the diameter: $\left( 0,0 \right)$ $2\text{nd}$ end of the diameter: $\left( -48,0 \right)$ Midpoint $=\left( \dfrac{0-48}{2},\dfrac{0+0}{2} \right)=\left( -24,0 \right)$ Therefore, the center of the circle has the coordinates: $\left( -24,0 \right)$ ${{\left( x+24 \right)}^{2}}+{{y}^{2}}=576$

 A student is trying to calculate the equation of a circle. If ${{\tan }^{-1}}\left( x \right)=60$ where $x$ is the length of the altitude dropped from the radius when it is at ${{60}^{\circ }}$ and the center of the circle is $\left( 3,5 \right)$, calculate the equation of the circle. 

 Calculate the inverse tangent of $x$ where it equals to ${{60}^{\circ }}$. Using the value of $x$, calculate the length of the radius by using the Pythagorean theorem. Put the coordinates of the center and the length of the radius in the equation of a circle.  

Center: $\left( 3,5 \right)$ ${{\tan }^{-1}}\left( x \right)=60 \\ \Rightarrow \tan 60=x \\ \Rightarrow x=\sqrt{3} $ Opposite side $=\sqrt{3}$ Adjacent side $=1$ ${{1}^{2}}+{{\left( \sqrt{3} \right)}^{2}}={{h}^{2}} \\ \Rightarrow 1+3={{h}^{2}} \\ \Rightarrow h=2 \\ $ ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}} \\ {{\left( x-3 \right)}^{2}}+{{\left( y-5 \right)}^{2}}=4 $

 The center of a circle is given by $\left( h,k \right)=\left( 3,4 \right)$. Calculate the area of the circle if the circle passes through the point $\left( 6,0 \right)$. 

 Calculate the distance between the center of the origin and the point $\left( 6,0 \right)$. Use the distance as the radius and apply the formula of the area of a circle $A=\pi {{r}^{2}}$.

$D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} \\ \Rightarrow D=\sqrt{{{\left( 6-3 \right)}^{2}}+{{\left( 0-4 \right)}^{2}}}=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( -4 \right)}^{2}}} \\ \Rightarrow D=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( -4 \right)}^{2}}}=\sqrt{9+16}=5 $ $A=\pi {{r}^{2}} \\ \Rightarrow A=\pi {{\left( 5 \right)}^{2}} \\ \Rightarrow A=25\pi $

Identify the angle formed between the two tangents that originate from the origin to the circle ${{\left( x-14 \right)}^{2}}+{{\left( y+2 \right)}^{2}}={{r}^{2}}$. If the length from the center of a circle to an arbitrary point on the circle is $10$ units. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7cb6be67-d02c-4cb9-a301-3d5d8df9d0084750472251066621209.png"/>

 Using the formula to calculate the distance between a line and a point, evaluate the value of $m$. Identify if the product of the slopes to identify if they are perpendicular or not. 

$y=mx+b \\ \Rightarrow y-mx=0 \\ $ Center of a circle: $\left( 14,-2 \right)$ $\dfrac{1\times \left( -2 \right)-m\left( 14 \right)}{\sqrt{{{1}^{2}}+{{m}^{2}}}}=10 \\ \Rightarrow \dfrac{-2-14m}{\sqrt{1+{{m}^{2}}}}=10 \\ \Rightarrow -2-14m=10\sqrt{1+{{m}^{2}}} \\ \Rightarrow m=-\dfrac{4}{3},\dfrac{3}{4} $ ${{m}_{1}}{{m}_{2}}=-\dfrac{4}{3}\times \dfrac{3}{4}=-1$ Since the product of the slopes is $-1$ The angle formed between the two tangents is ${{90}^{\circ }}$

 A student is trying to calculate normal to the equation of a circle ${{x}^{2}}+{{y}^{2}}=576$, if the normal is at the point $\left( \dfrac{1}{8\sqrt{2}},\dfrac{1}{16\sqrt{2}} \right)$

 Using the equation of a normal to the circle ${{x}^{2}}+{{y}^{2}}=576$ at a point $\left( {{a}_{1}},{{b}_{1}} \right)$, put the value of the point in $\dfrac{x}{{{a}_{1}}}-\dfrac{y}{{{b}_{1}}}=0$ and simplify it to get the equation of the normal.

Equation of a normal to the circle at the point $\left( {{a}_{1}},{{b}_{1}} \right)$ is $\dfrac{x}{{{a}_{1}}}-\dfrac{y}{{{b}_{1}}}=0$ Point: $\left( \dfrac{1}{8\sqrt{2}},\dfrac{1}{16\sqrt{2}} \right)$ Normal: $\dfrac{x}{\dfrac{1}{8\sqrt{2}}}-\dfrac{y}{\dfrac{1}{16\sqrt{2}}}=0$ $\Rightarrow 8\sqrt{2}x-16\sqrt{2}y=0 \\ \Rightarrow 8\sqrt{2}\left( x-2y \right)=0 \\ \Rightarrow x-2y=0$