Centre of circle is at origin …

CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Ninad Joshi 6 years, 11 months ago
- 2 answers
Ram Kushwah 6 years, 11 months ago
If O(,0,0) is origin A(3,4) is a pint on circumference then
r=OB{ (3-0)2 +(4-0)2 }1/2
r=5
{tex}\begin{array}{l}Area=\mathrm{πr}^2=3.14\times5\times5=78.5\;\mathrm{sq}\;\mathrm{unit}\\\mathrm{circumference}=2\mathrm{πr}=2\times3.14\times5=31.4\\\end{array}{/tex}
Related Questions
Posted by Parinith Gowda Ms 3 months, 3 weeks ago
- 0 answers
Posted by Kanika . 1 month, 1 week ago
- 1 answers
Posted by Sahil Sahil 1 year, 4 months ago
- 2 answers
Posted by Vanshika Bhatnagar 1 year, 4 months ago
- 2 answers
Posted by Lakshay Kumar 1 year, 1 month ago
- 0 answers
Posted by Parinith Gowda Ms 3 months, 3 weeks ago
- 1 answers
Posted by Hari Anand 6 months, 2 weeks ago
- 0 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
myCBSEguide
Anushka ??? 6 years, 11 months ago
1Thank You