LIVE Course for free

Rated by 1 million+ students
Get app now
Class 8 Foundation Course
Class 9 Foundation Course
Class 10 Foundation Course
Class 11 Foundation Course
Class 12 Foundation Course
JEE COURSE
NEET COURSE
0 votes
24 views
in Circles by (54.3k points)
closed by
Find radius of the circle: 3x2+ 3y2- 6x+ 12y- 13= 0
1. \(\sqrt{\frac{28}{3}}\) units
2. \(\sqrt{\frac{13}{3}}\) units
3. \({\frac{\sqrt28}{3}}\) units
4. \({\frac{\sqrt13}{2}}\) units

1 Answer

0 votes
by (30.0k points)
selected by
 
Best answer
Correct Answer - Option 1 : \(\sqrt{\frac{28}{3}}\) units

Concept: 

General form of the equation of a circle, x2 + y2 + 2gx + 2fy + c = 0 

Radius = \(\rm\sqrt{g^{2}+f^{2}-c}\)  

Calculation: 

The given equation of circle is , 3x2+ 3y2- 6x+ 12y- 13= 0 

⇒ x2+ y2- 2x + 4y - \(\frac{13}{3}\) = 0         ....(i)

On compare with standard equation of circle x2 + y2 + 2gx + 2fy + c = 0  

where , g = 1 , f = 2 and c = - \(\frac{13}{3}\)  

We know that , radius of circle = \(\rm\sqrt{g^{2}+f^{2}-c}\) 

⇒ radius = \(\sqrt{1^{2}+2^{2}-\left ( -\frac{13}{3} \right )}\) 

radius\(\sqrt{\frac{28}{3}}\) units. 

The correct option is 1.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...