Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • What is the equation of the circle that passes through the points <img alt="(1,1),(2,3), \text { and }(-1,2) ?" src="https://entrancecorner.oncodecogs.com/gif.latex?%281%2C1%29%2C%282%2C3%29%2

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

Option: 1

(x-2)^2+(y-1)^2=1


Option: 2

(x-1)^2+(y-2)^2=5


Option: 3

(x-2)^2+(y-1)^2=5


Option: 4

(x-1)^2+(y-3)^2=5


Answers (1)

best_answer

To find the equation of the circle that passes through the points (1,1),(2,3) \text {, and }(-1,2) \text {, } we can use the fact that the center of the circle lies at the intersection of the perpendicular bisectors of any two chords of the circle.

Let's first find the midpoint and the slope of the line segment joining the points (1,1)\, and\, (2,3) .

Midpoint of (1,1) \text { and }(2,3)=\left(\frac{1+2}{2}, \frac{1+3}{2}\right)=\left(\frac{3}{2}, 2\right)

The slope of the line segment joining (1,1) \text { and }(2,3)=\left(\frac{3-1}{2-1}\right)=2

The equation of the perpendicular bisector of the line segment joining \left ( 1,1 \right )\, and\, \left ( 2,3 \right ).can be written as:

(y-2)=\frac{-1}{2}\left(x-\frac{3}{2}\right)

Simplifying, we get:

x+2 y=\frac{11}{2}

Similarly, let's find the midpoint and slope of the line segment joining the points

\left ( 2,3 \right )and\left ( -1,2 \right ).

Midpoint of (2,3) \text { and }(-1,2)=\left(\frac{2-1}{2}, \frac{3+2}{2}\right)=\left(\frac{1}{2}, \frac{5}{2}\right)

The slope of the line segment joining (2,3) \text { and }(-1,2)=\left(\frac{2-3}{-1-2}\right)=\left(\frac{-1}{3}\right)

The equation of the perpendicular bisector of the line segment joining \left ( 2,3 \right )and\left ( -1,2 \right ) can be written as:

\mathrm{\left ( \frac{y-5}{2} \right )=3\left ( x-\frac{1}{2} \right )}

Simplifying, we get:

\mathrm{3x-y=\frac{7}{2}}

The intersection point of these two perpendicular bisectors is the center of the circle.

Solving for x and y, we get: \mathrm{x=2\, and\, y=1}

So, the centre of the circle is (2,1) and the radius of the circle can be found by calculating the distance between any of the three given points and the centre of the circle.

Let's use the point (1,1):

\mathrm{\text { radius }=\sqrt{(2-1)^2+(1-1)^2}=1}

Therefore, the equation of the circle is:

\mathrm{(x-2)^2+(y-1)^2=1^2}

Simplifying, we get:\mathrm{(x-2)^2+(y-1)^2=1}

 

 

Posted by

HARSH KANKARIA

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 registration ends tomorrow; changes introduced in exam this year
anam-khan

BTech Admission 2025 Live: JEE Mains registrations closing tomorrow; VITEEE, MET exam dates out
anam-khan

JEE Main 2025 Live: NTA to close JEE registration on Friday; apply at jeemain.nta.nic.in, correction window

Latest Question