# Find the equation of the circle passing through the intersection of the lines \[3x+y=4\] and \[x-3y+2=0\] and concentric with the circle \[2\left( {{x}^{2}}+{{y}^{2}} \right)-3x+8y-1=0\].

Last updated date: 26th Mar 2023

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Answer

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Hint: Two concentric circles have same center and distance from center to any point on circumference is radius

Given lines, \[{{L}_{1}}=3x+y=4....\left( i \right)\]

\[{{L}_{2}}=x-3y+2=0....\left( ii \right)\]

We have to find the point of intersection of \[{{ L }_{1}}\]and \[{{L}_{2}}\], we will solve equation \[\left( i \right)\]and \[\left( ii \right)\]together.

Taking equation\[\left( i \right)\], \[3x+y=4\]

We get, \[y=4-3x....\left( iii \right)\]

Putting value of \[y\] in equation \[\left( ii \right)\]

\[x-3y+2=0\]

\[x-3\left[ 4-3x \right]+2=0\]

\[x-12+9x+2=0\]

\[10x=10\]

Therefore, we get

Putting value of \[x\] in equation \[\left( iii \right)\]to find the value of \[y\]

\[y=4-3x\]

\[=4-3\left( 1 \right)\]

We get \[y=1\]

Therefore circle \[{{C}_{2}}\]passes through\[\left( x,y \right)=\left( 1,1 \right)\].

Now circle \[{{C}_{2}}\]is concentric with circle:

\[{{C}_{1}}=2\left( {{x}^{2}}+{{y}^{2}} \right)-3x+8y-1=0\]

Dividing the equation by \[2\],

We get \[{{C}_{1}}=\left( {{x}^{2}}+{{y}^{2}} \right)-\dfrac{3x}{2}+4y-\dfrac{1}{2}=0....\left( iv \right)\]

The general equation of circle is,

\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]

Now, we will compare general equation of circle with equation\[\left( iv \right)\].

We get, \[2g=-\dfrac{3}{2}\]

\[g=\dfrac{-3}{4}\]

\[2f=4\]

\[f=2\]

We know that center of circle is \[\left( -g,-f \right)\].

Therefore, \[\text{centre}=\left( \dfrac{3}{4},-2 \right)\]

Since, given circle is concentric with circle to be found \[\left( {{C}_{2}} \right)\], both would have same center.

Now, the new circle \[{{C}_{2}}\]has center at \[C\left( \dfrac{3}{4},-2 \right)\]and it passes through intersection of line \[A=\left( 1,1 \right)\].

Distance of point \[A\left( 1,1 \right)\]to \[C\left( \dfrac{3}{4},-2 \right)\]is radius.

By distance formula,

\[r=AC=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]

\[r=AC=\sqrt{{{\left( 1-\dfrac{3}{4} \right)}^{2}}+{{\left( 1+2 \right)}^{2}}}\]

\[=\sqrt{\dfrac{1}{16}+9}\]

We get radius \[=\sqrt{\dfrac{145}{16}}=\dfrac{\sqrt{145}}{4}\text{ units}\]

Now we know that, equation of circle with center \[\left( a,b \right)\]and radius \[r\]

\[={{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}\]

Here, we found the center \[\left( \dfrac{3}{4},-2 \right)\]and radius\[=\dfrac{\sqrt{145}}{4}\text{ units}\]

We get, equation of circle \[={{\left( x-\dfrac{3}{4} \right)}^{2}}+{{\left[ y-\left( -2 \right) \right]}^{2}}=\dfrac{145}{16}\]

\[=16{{x}^{2}}+16{{y}^{2}}-24x+64y=72\]

Therefore, final equation of circle is:

\[2\left( {{x}^{2}}+{{y}^{2}} \right)-3x+8y=9\]

Note:

Elimination method is also suitable to find the intersection of line. Students can also use the method of family of circles passing through a point and having a center with constraint.

Given lines, \[{{L}_{1}}=3x+y=4....\left( i \right)\]

\[{{L}_{2}}=x-3y+2=0....\left( ii \right)\]

We have to find the point of intersection of \[{{ L }_{1}}\]and \[{{L}_{2}}\], we will solve equation \[\left( i \right)\]and \[\left( ii \right)\]together.

Taking equation\[\left( i \right)\], \[3x+y=4\]

We get, \[y=4-3x....\left( iii \right)\]

Putting value of \[y\] in equation \[\left( ii \right)\]

\[x-3y+2=0\]

\[x-3\left[ 4-3x \right]+2=0\]

\[x-12+9x+2=0\]

\[10x=10\]

Therefore, we get

Putting value of \[x\] in equation \[\left( iii \right)\]to find the value of \[y\]

\[y=4-3x\]

\[=4-3\left( 1 \right)\]

We get \[y=1\]

Therefore circle \[{{C}_{2}}\]passes through\[\left( x,y \right)=\left( 1,1 \right)\].

Now circle \[{{C}_{2}}\]is concentric with circle:

\[{{C}_{1}}=2\left( {{x}^{2}}+{{y}^{2}} \right)-3x+8y-1=0\]

Dividing the equation by \[2\],

We get \[{{C}_{1}}=\left( {{x}^{2}}+{{y}^{2}} \right)-\dfrac{3x}{2}+4y-\dfrac{1}{2}=0....\left( iv \right)\]

The general equation of circle is,

\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]

Now, we will compare general equation of circle with equation\[\left( iv \right)\].

We get, \[2g=-\dfrac{3}{2}\]

\[g=\dfrac{-3}{4}\]

\[2f=4\]

\[f=2\]

We know that center of circle is \[\left( -g,-f \right)\].

Therefore, \[\text{centre}=\left( \dfrac{3}{4},-2 \right)\]

Since, given circle is concentric with circle to be found \[\left( {{C}_{2}} \right)\], both would have same center.

Now, the new circle \[{{C}_{2}}\]has center at \[C\left( \dfrac{3}{4},-2 \right)\]and it passes through intersection of line \[A=\left( 1,1 \right)\].

Distance of point \[A\left( 1,1 \right)\]to \[C\left( \dfrac{3}{4},-2 \right)\]is radius.

By distance formula,

\[r=AC=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]

\[r=AC=\sqrt{{{\left( 1-\dfrac{3}{4} \right)}^{2}}+{{\left( 1+2 \right)}^{2}}}\]

\[=\sqrt{\dfrac{1}{16}+9}\]

We get radius \[=\sqrt{\dfrac{145}{16}}=\dfrac{\sqrt{145}}{4}\text{ units}\]

Now we know that, equation of circle with center \[\left( a,b \right)\]and radius \[r\]

\[={{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}\]

Here, we found the center \[\left( \dfrac{3}{4},-2 \right)\]and radius\[=\dfrac{\sqrt{145}}{4}\text{ units}\]

We get, equation of circle \[={{\left( x-\dfrac{3}{4} \right)}^{2}}+{{\left[ y-\left( -2 \right) \right]}^{2}}=\dfrac{145}{16}\]

\[=16{{x}^{2}}+16{{y}^{2}}-24x+64y=72\]

Therefore, final equation of circle is:

\[2\left( {{x}^{2}}+{{y}^{2}} \right)-3x+8y=9\]

Note:

Elimination method is also suitable to find the intersection of line. Students can also use the method of family of circles passing through a point and having a center with constraint.

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