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What is the equation of the circle that passes through the points (1,1),(2,3), \text { and }(-1,2) ?

Option: 1

\left(x+\frac{2}{7}\right)^2+\left(y-\frac{5}{7}\right)^2=\left(\frac{3}{7}\right)^2 \\


Option: 2

\left(x+\frac{5}{7}\right)^2+\left(y-\frac{17}{7}\right)^2=\left(\frac{6}{7}\right)^2 \\


Option: 3

\left(x-\frac{2}{7}\right)^2+\left(y-\frac{5}{7}\right)^2=\left(\frac{3}{7}\right)^2 \\


Option: 4

\left(x-\frac{5}{7}\right)^2+\left(y-\frac{17}{7}\right)^2=\left(\frac{6}{7}\right)^2


Answers (1)

best_answer

To find the equation of the circle, we need to find the centre and radius of the circle.

We can use the fact that the perpendicular bisectors of the chords of a circle pass through the centre of the circle.

So, we can find the equations of the perpendicular bisectors of the chords that connect the three given points, and then find their point of intersection. This point of intersection will be the center of the circle.

Let's find the equation of the perpendicular bisector of the chord that connects \left ( 1,1 \right )and\left ( 2,3 \right ).

The midpoint of the chord is \left[\frac{1+2}{2}, \frac{1+3}{2}\right]=(1.5,2)

The slope of the chord is \frac{(3-1)}{(2-1)}=2

The slope of the perpendicular bisector is \frac{-1}{2} (negative reciprocal of 2)

The equation of the perpendicular bisector is:

\begin{aligned} & y-2=\frac{-1}{2}(x-1.5) \\ & y=\frac{-1}{2} x+3 \end{aligned}

Similarly, we can find the equations of the perpendicular bisectors of the chords that connect \left ( 2,3 \right )and\left ( -1,2 \right ) and \left ( -1,2 \right )and\left ( 1,1 \right ).

The equations are

 

\begin{aligned} & y=\frac{2}{3} x+\frac{8}{3} \\ & y=\frac{-1}{3} x+\frac{7}{3} \end{aligned}

To find the centre of the circle, we need to find the point of intersection of any two perpendicular bisectors. Let's take the first and second equations:

\begin{aligned} & \frac{-1}{2} x+3=\frac{2}{3} x+\frac{8}{3} \\ & \frac{7}{6} x=\frac{5}{3} \\ & x=\frac{5}{7} \end{aligned}

Substituting x=\frac{5}{7} in the first equation, we get:

\begin{aligned} & y=\frac{-1}{2} \frac{5}{7}+3 \\ & y=\frac{17}{7} \end{aligned}

Therefore, the center of the circle is \frac{5}{7},\frac{17}{7}To find the radius of the circle, we can use the distance between the centre and any of the three given points. Let's use (1,1):

\begin{aligned} r & =\sqrt{(1-5 / 7)^2+(1-17 / 7)^2} \\ r & =\sqrt{\frac{20}{49}+\frac{16}{49}} \\ r & =\sqrt{\frac{36}{49}} \\ r & =\frac{6}{7} \end{aligned}

Therefore, the equation of the circle is:

\left(x-\frac{5}{7}\right)^2+\left(y-\frac{17}{7}\right)^2=\left(\frac{6}{7}\right)^2

 

 

Posted by

Irshad Anwar

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