Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

The centroid of a circle is \[\left( {2, - 3} \right)\] and circumference is \[10\pi \] .Then the equation of the circle is
A.\[{x^2} + {y^2} + 4x + 6y + 12 = 0\]
B.\[{x^2} + {y^2} - 4x + 6y + 12 = 0\]
C.\[{x^2} + {y^2} - 4x + 6y - 12 = 0\]
D.\[{x^2} + {y^2} - 4x - 6y - 12 = 0\]

seo-qna
Last updated date: 14th Jun 2024
Total views: 402.3k
Views today: 5.02k
Answer
VerifiedVerified
402.3k+ views
Hint: We are given with the center of the circle and its circumference. From this we will find the radius of the circle using the formula \[2\pi r\] . Then using the general equation of circle \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\] and putting the value of center of circle we will get equation of circle.

Complete step-by-step answer:
Given that, circumference of a circle is \[10\pi \]
 \[ \Rightarrow 10\pi = 2\pi r\]
Cancelling \[\pi \] from both sides,
\[ \Rightarrow r = 5unit.\]
Now we know that the general form of the equation is \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\].
Center of the circle is \[\left( {h,k} \right) = \left( {2, - 3} \right)\] and radius \[r = 5\].
Putting these values in the equation above
\[ \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - \left( { - 3} \right)} \right)^2} = {5^2}\]
Performing the expansions using the identity
\[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\] and \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]
\[
   \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - \left( { - 3} \right)} \right)^2} = {5^2} \\
   \Rightarrow {x^2} - 4x + 4 + \left( {{y^2} + 6y + 9} \right) = 25 \\
   \Rightarrow {x^2} + {y^2} - 4x + 6y + 4 + 9 = 25 \\
   \Rightarrow {x^2} + {y^2} - 4x + 6y + 13 = 25 \\
   \Rightarrow {x^2} + {y^2} - 4x + 6y = 25 - 13 \\
   \Rightarrow {x^2} + {y^2} - 4x + 6y = 12 \\
\]
And this is the equation of the circle \[{x^2} + {y^2} - 4x + 6y = 12\].
Hence option B is correct.

Note: We are given with four options here with slight difference in the signs only. So be careful when you expand the brackets and add or subtract the terms. Because a minor negligence will make your answer wrong.