Answer

Verified

478.5k+ views

Hint: We will put the given points in the general equation of the circle and find the values of \[\left( g,f \right)\] and \[c\] and then form the required equation of the circle.

Given that \[\left( 1,1 \right),\left( 2,-1 \right)\] and \[\left( 3,2 \right)\] are three non-collinear points. All of these non-collinear points lie on the circle.

We have to find the equation of the given circle.

We know that the general equation of the circle is

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

All of these non-collinear points lie on the circle.

Putting the values of point \[A\left( 1,1 \right)\] in the general equation of the circle, we get

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

\[\Rightarrow {{1}^{2}}+{{1}^{2}}+2g\left( 1 \right)+2f\left( 1 \right)+c=0\]

\[\Rightarrow 1+1+2g+2f+c=0\]

\[\Rightarrow 2+2g+2f+c=0\]

\[\Rightarrow 2g+2f+c=-2.....\left( i \right)\]

Putting the value of point \[B\left( 2,-1 \right)\] in the general equation of the circle, we get,

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

\[\Rightarrow {{2}^{2}}+{{\left( -1 \right)}^{2}}+2g\left( 2 \right)+2f\left( -1 \right)+c=0\]

\[\Rightarrow 4+1+4g-2f+c=0\]

\[\Rightarrow 4g-2f=-c-5.....\left( ii \right)\]

Putting the value of point \[C\left( 3,2 \right)\] in the general equation of the circle, we get,

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

\[\Rightarrow {{3}^{2}}+{{\left( 2 \right)}^{2}}+2g\left( 3 \right)+2f\left( 2 \right)+c=0\]

\[\Rightarrow 9+4+6g+4f+c=0\]

\[\Rightarrow 6g+4f=-c-13.....\left( iii \right)\]

Now, we have three equations and three unknowns i.e. \[g,f\] and \[c\] respectively.

Solving equation \[\left( i \right)\] and \[\left( ii \right)\], we get

\[\begin{align}

& 2g+2f=-c-2 \\

& \underline{4g-2f=-c-5} \\

& 6g+0=-2c-7 \\

\end{align}\]

Putting the value of \[6g\] in equation \[\left( iii \right)\],

\[\Rightarrow 6g+4f=-c-13\]

\[\Rightarrow -2c-7+4f=-c-13\]

\[\Rightarrow 4f=c-6\]

\[\Rightarrow c=4f+6\]

Now, putting the value of \[c\] in the equation \[\left( i \right)\], we get

\[\Rightarrow 2g+2f=-4f-6-2\]

\[\Rightarrow 2g+2f=-4f-8\]

\[\Rightarrow 2g+6f=-8.....\left( v \right)\]

Putting the value of \['c'\] from the equation \[\left( iv \right)\] in the equation \[\left( ii \right)\], we get

\[\Rightarrow 4g-2f=-4f-6-5\]

\[\Rightarrow 4g+2f=-11.....\left( vi \right)\]

Solving equation \[\left( v \right)\] and \[\left( vi \right)\], we get

\[\begin{align}

& 4g+12f=-16 \\

& 4g+2f=-11 \\

& \underline{-\text{ - + }} \\

& 0+10f=-5 \\

\end{align}\]

\[f=-\dfrac{1}{2}\]

Put the value of \[f\]in the equation \[\left( v \right)\], we get

\[2g+6\times \left( -\dfrac{1}{2} \right)=-8\]

\[g=\dfrac{-8+3}{2}\]

\[g=\dfrac{-5}{2}\]

Similarly, putting the value of \[f\] in the equation \[\left( iv \right)\], we get

\[\Rightarrow c=4f+6\]

\[\Rightarrow c=4\times \left( \dfrac{-1}{2} \right)+6\]

\[\Rightarrow c=4\]

Now, we have all the required points to form an equation of the circle.

Thus, by putting the values in the general equation of the circle, we get

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

\[\Rightarrow {{x}^{2}}+{{y}^{2}}+2\left( \dfrac{-5}{2} \right)x+2\left( \dfrac{-1}{2} \right)y+4=0\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}-5x-y+4=0\]

Note: Alternative method:

You can solve this question by using the distance formula

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

\[d\] is the distance between two points.

By putting the values of the given point in the distance formula, we can find the coordinates of the circle and its radius and finally the required equation of the circle.

Given that \[\left( 1,1 \right),\left( 2,-1 \right)\] and \[\left( 3,2 \right)\] are three non-collinear points. All of these non-collinear points lie on the circle.

We have to find the equation of the given circle.

We know that the general equation of the circle is

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

All of these non-collinear points lie on the circle.

Putting the values of point \[A\left( 1,1 \right)\] in the general equation of the circle, we get

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

\[\Rightarrow {{1}^{2}}+{{1}^{2}}+2g\left( 1 \right)+2f\left( 1 \right)+c=0\]

\[\Rightarrow 1+1+2g+2f+c=0\]

\[\Rightarrow 2+2g+2f+c=0\]

\[\Rightarrow 2g+2f+c=-2.....\left( i \right)\]

Putting the value of point \[B\left( 2,-1 \right)\] in the general equation of the circle, we get,

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

\[\Rightarrow {{2}^{2}}+{{\left( -1 \right)}^{2}}+2g\left( 2 \right)+2f\left( -1 \right)+c=0\]

\[\Rightarrow 4+1+4g-2f+c=0\]

\[\Rightarrow 4g-2f=-c-5.....\left( ii \right)\]

Putting the value of point \[C\left( 3,2 \right)\] in the general equation of the circle, we get,

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

\[\Rightarrow {{3}^{2}}+{{\left( 2 \right)}^{2}}+2g\left( 3 \right)+2f\left( 2 \right)+c=0\]

\[\Rightarrow 9+4+6g+4f+c=0\]

\[\Rightarrow 6g+4f=-c-13.....\left( iii \right)\]

Now, we have three equations and three unknowns i.e. \[g,f\] and \[c\] respectively.

Solving equation \[\left( i \right)\] and \[\left( ii \right)\], we get

\[\begin{align}

& 2g+2f=-c-2 \\

& \underline{4g-2f=-c-5} \\

& 6g+0=-2c-7 \\

\end{align}\]

Putting the value of \[6g\] in equation \[\left( iii \right)\],

\[\Rightarrow 6g+4f=-c-13\]

\[\Rightarrow -2c-7+4f=-c-13\]

\[\Rightarrow 4f=c-6\]

\[\Rightarrow c=4f+6\]

Now, putting the value of \[c\] in the equation \[\left( i \right)\], we get

\[\Rightarrow 2g+2f=-4f-6-2\]

\[\Rightarrow 2g+2f=-4f-8\]

\[\Rightarrow 2g+6f=-8.....\left( v \right)\]

Putting the value of \['c'\] from the equation \[\left( iv \right)\] in the equation \[\left( ii \right)\], we get

\[\Rightarrow 4g-2f=-4f-6-5\]

\[\Rightarrow 4g+2f=-11.....\left( vi \right)\]

Solving equation \[\left( v \right)\] and \[\left( vi \right)\], we get

\[\begin{align}

& 4g+12f=-16 \\

& 4g+2f=-11 \\

& \underline{-\text{ - + }} \\

& 0+10f=-5 \\

\end{align}\]

\[f=-\dfrac{1}{2}\]

Put the value of \[f\]in the equation \[\left( v \right)\], we get

\[2g+6\times \left( -\dfrac{1}{2} \right)=-8\]

\[g=\dfrac{-8+3}{2}\]

\[g=\dfrac{-5}{2}\]

Similarly, putting the value of \[f\] in the equation \[\left( iv \right)\], we get

\[\Rightarrow c=4f+6\]

\[\Rightarrow c=4\times \left( \dfrac{-1}{2} \right)+6\]

\[\Rightarrow c=4\]

Now, we have all the required points to form an equation of the circle.

Thus, by putting the values in the general equation of the circle, we get

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

\[\Rightarrow {{x}^{2}}+{{y}^{2}}+2\left( \dfrac{-5}{2} \right)x+2\left( \dfrac{-1}{2} \right)y+4=0\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}-5x-y+4=0\]

Note: Alternative method:

You can solve this question by using the distance formula

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

\[d\] is the distance between two points.

By putting the values of the given point in the distance formula, we can find the coordinates of the circle and its radius and finally the required equation of the circle.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

How do you graph the function fx 4x class 9 maths CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths