Find the equation of the circle …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Related Questions
Posted by Sreenidhil Kp 3 years, 11 months ago
- 0 answers
Posted by Derpak Yadav 3 years, 2 months ago
- 0 answers
Posted by Harsh Prakash 1 year, 9 months ago
- 2 answers
Posted by Nagendra Gundappanavar 3 years, 9 months ago
- 1 answers
Posted by Namrata Pasi 2 years, 5 months ago
- 0 answers
Posted by Sreenidhil Kp 3 years, 11 months ago
- 1 answers
Posted by Rahul Roy 3 years, 3 months ago
- 1 answers
Posted by Anuj Kumar Yadav 2 years, 2 months ago
- 2 answers
Posted by Masuka Addwin 3 years, 8 months ago
- 2 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Yogita Ingle 3 years, 11 months ago
Let the equation of circle be
x2+y2+2gx+2fy+c=0 .....(1)
where (−g,−f) is center and g2+f2−c is the radius.
Since, the circle passes through (2,3), so it satisfies eqn (1)
⇒4+9+4g+6f+c=0
⇒4g+6f+c=−13 .....(2)
Since, the circle passes through (−1,1), so it satisfies eqn (1)
⇒1+1−2g+2f+c=0
⇒−2g+2f+c=−2 .....(3)
Subtracting eqn (3) from eqn (2), we get
6g+4f=−11 ....(4)
Given center lies on the line x−3y−11=0
Since, center (−g,−f) satisfies this equation.
⇒−g+3f=11 ....(5)
Solving eqn (4) and (5), we get
g= −7/2 ,f= 5/2
Put this value in (2), we get
c=−14
Substituting these values in (1),
x2+y2−7x+5y−14=0
which is the equation of required circle.
0Thank You