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

Circles Quiz 2

mathematics

Find the equation of the circle with the center at $\left( \left( a+b \right),\left( a-b \right) \right)$  and the radius as $\sqrt{{{a}^{2}}+{{b}^{2}}}$ .

  Put the value of center and radius in the general equation of the circle with center $\left( h,k \right)$ and radius $r$ is given by ${{\left( x-h \right)}^{2}}-{{\left( y-k \right)}^{2}}={{r}^{2}}$.  

$ \text{The equation of the circle with center }\left( h,k \right)\text{ and radius }r\text{ is given as} \\  \Rightarrow {{\left( x-h \right)}^{2}}-{{\left( y-k \right)}^{2}}={{r}^{2}} \\ \text{Putting the values of center and radius in the equation,} \\ \left\{ x-{{\left( a+b \right)}^{2}} \right\}+\left\{ y-{{\left( a-b \right)}^{2}} \right\}={{\left( \sqrt{{{a}^{2}}+{{b}^{2}}} \right)}^{2}} \\ \Rightarrow {{x}^{2}}+{{\left( a+b \right)}^{2}}-2ax-2xb+{{y}^{2}}+{{\left( a-b \right)}^{2}}-2ay+2ab={{a}^{2}}+{{b}^{2}} \\ \Rightarrow {{x}^{2}}+{{y}^{2}}-2ax-2bx-2ay+2by+{{a}^{2}}+{{b}^{2}}=0$

  Determine the center of the circle and radius having the equation as $2{{x}^{2}}-2cx+2{{y}^{2}}-2dy=0$ 

 Simplify the equation and use the method of completing the whole square and then compare with the standard equation of the circle to get the required answer.

 $ \Rightarrow 2{{x}^{2}}-2cx+2{{y}^{2}}-2dy=0 \\ \Rightarrow \left( {{x}^{2}}-cx \right)+\left( {{y}^{2}}-dy \right)=0 \\ \Rightarrow \left( {{x}^{2}}-cx-\dfrac{{{c}^{2}}}{4}+\dfrac{{{c}^{2}}}{4} \right)+\left( {{y}^{2}}-dy-\dfrac{{{d}^{2}}}{4}+\dfrac{{{d}^{2}}}{4} \right)=0 \\ \Rightarrow {{\left( x-\dfrac{c}{2} \right)}^{2}}+{{\left( y-\dfrac{d}{2} \right)}^{2}}-\dfrac{{{c}^{2}}}{4}-\dfrac{{{d}^{2}}}{4}=0 \\ \Rightarrow {{\left( x-\dfrac{c}{2} \right)}^{2}}+{{\left( y-\dfrac{d}{2} \right)}^{2}}=\dfrac{{{c}^{2}}}{4}+\dfrac{{{d}^{2}}}{4} \\ \Rightarrow {{\left( x-\dfrac{c}{2} \right)}^{2}}+{{\left( y-\dfrac{d}{2} \right)}^{2}}={{\left( \dfrac{1}{2}\sqrt{{{c}^{2}}+{{d}^{2}}} \right)}^{2}} \\$<br>Comparing the equation with the standard equation of the circle the center is at $\left( \dfrac{c}{2},\dfrac{d}{2} \right)$ and the radius is  $\dfrac{1}{2}\sqrt{{{c}^{2}}+{{d}^{2}}}$ units

 Find the area of the circle with equation  ${{x}^{2}}+{{y}^{2}}=2\pi \left( \alpha +\beta  \right)$ .

 Calculate the radius by solving the equation then use the formula $\pi {{r}^{2}}$ to calculate the area.

 $ {{x}^{2}}+{{y}^{2}}=2\pi \left( \alpha x+\beta y \right) \\ \Rightarrow {{x}^{2}}+{{y}^{2}}-2\pi \left( \alpha x+\beta y \right)=0 \\ \Rightarrow {{x}^{2}}+{{y}^{2}}-2\pi \alpha x-2\pi \beta y=0 \\ \Rightarrow \left( {{x}^{2}}-2\pi \alpha x \right)+\left( {{y}^{2}}-2\pi \beta y \right)=0 \\ \Rightarrow \left( {{x}^{2}}-2\pi \alpha x+{{\pi }^{2}}{{\alpha }^{2}}-{{\pi }^{2}}{{\alpha }^{2}} \right)+\left( {{y}^{2}}-2\pi \beta y+{{\pi }^{2}}{{\beta }^{2}}-{{\pi }^{2}}{{\beta }^{2}} \right)=0 \\ \Rightarrow {{\left( x-\pi \alpha  \right)}^{2}}+{{\left( y-\pi \beta  \right)}^{2}}={{\pi }^{2}}\left( {{\alpha }^{2}}+{{\beta }^{2}} \right) \\ \text{The radius of the circle is }\pi \sqrt{\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)} \\ \text{Now, the area of the circle is,} \\ \Rightarrow A=\pi {{r}^{2}}={{\pi }^{3}}\left( {{\alpha }^{2}}+{{\beta }^{2}} \right) $

 What is the position of point $\left( 10,6 \right)$ with reference to the circle with the equation ${{x}^{2}}+{{y}^{2}}-6x-10y=0$ 

 Substitute the given point in the circle equation and check the vale. If it is greater than zero then the given point is outside the circle.

 $ \text{Substitute the given point in the equation of circle,} \\ \Rightarrow LHS={{x}^{2}}+{{y}^{2}}-6x-10y \\ \Rightarrow LHS={{10}^{2}}+{{6}^{2}}-6\left( 10 \right)-10\left( 6 \right) \\ \Rightarrow LHS=100+36-60-60 \\ \Rightarrow LHS=16>0 \\ \text{So, the given point lie outside the circle}\text{.} $

 Raju wants to find the equation of a line passing through the centre of the circle having the equation ${{x}^{2}}+{{y}^{2}}+8x-20y+16=0$. The required line passes through a point on the circle such that the equation of tangent at that point is given as $4x+3y=64$. Find the required equation.

  Convert the given circle equation to standard form to find the center. Then find the slope of the required equation which is perpendicular to the tangent and then find the required equation using the slope point form of a line.

 $ {{x}^{2}}+{{y}^{2}}+8x-20y+16=0 \\ \Rightarrow {{\left( x+4 \right)}^{2}}+{{\left( y-10 \right)}^{2}}=16+100-16 \\ \Rightarrow {{\left( x+4 \right)}^{2}}+{{\left( y-10 \right)}^{2}}={{10}^{2}} \\ \text{Centre of the circle is }\left( -4,10 \right) \\ \text{Given the tangent equation as }4x+3y=64 \\ \Rightarrow \text{Slope, }m=\dfrac{-4}{3} \\ \text{The line passing through center is perpendicular to tangent at point on circle}\text{. } \\ \text{So, the slope of required equation is} \\ \Rightarrow M=\dfrac{3}{4} \\ \text{Now, the required equation is} \\ \Rightarrow y-10=\dfrac{3}{4}\left( x+4 \right) \\ \Rightarrow 4y-40=3x+12 \\ \Rightarrow 3x-4y+52=0$