Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
61 views
in Circles by (115k points)
closed by
Find the equation of a circle with centre at (2, - 3) and radius 5 units.
1. x2 + y2 - 4x + 6y - 12 = 0
2. x2 + y2 - 4x - 6y - 12 = 0
3. x2 + y2 + 10x + 2y + 22 = 0
4. None of these
5. x2 + y2 + 10x + 2y - 22 = 0

1 Answer

0 votes
by (114k points)
selected by
 
Best answer
Correct Answer - Option 1 : x2 + y2 - 4x + 6y - 12 = 0

Concept:

The equation of circle with centre at (h, k) and radius 'r' is 

(x - h)+ (y - k)= r2

Calculation:

We know that, the equation of circle with centre at (h, k) and radius 'r' is 

(x - h)+ (y - k)= r2

Here, centre (h, k) = (2, - 3) and radius r = 5 units.

Hence, the equation of a circle with centre at (2, - 3) and radius 5 units is 

(x - 2)+ (y + 3)= 52

⇒ \(\rm x^2 - 4x + 4 + y^2 +6y +9 = 25\)

⇒ \(\rm x^2 + y^2 - 4x + 6y - 12 = 0 \)

Hence, the equation of a circle with centre at (2, - 3) and radius 5 units is \(\rm x^2 + y^2 - 4x + 6y - 12 = 0 \).

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

...