The centre of a circle is 2- -3 and the circumference is 10- Then the equation of the circle is

Centre (2, – 3) Circumference = 10π ⇒ 2π r = 10 π ⇒ r = 5. From (x h)^2 + (y k)^2 = r^2, (x 2)^2 + (y+3)^2 = 5^2 ⇒ x^2 + y^2 4x + 6y + 13 = 25 ⇒ x^2 + y^2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Equation of Circle with (h,k) as Center

The centre of...

Question

The centre of a circle is (2, –3) and the circumference is $10 π .$ Then the equation of the circle is

x^{2} + y^{2} + 4 x + 6 y + 12 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} - 4 x + 6 y + 12 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} - 4 x + 6 y - 12 = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x^{2} + y^{2} - 4 x - 6 y - 12 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$
Centre (2, – 3)
Circumference = $10 π \Rightarrow 2 π r = 10 π \Rightarrow r = 5.$
From $(x - h)^{2} + (y - k)^{2} = r^{2},$
$(x - 2)^{2} + (y + 3)^{2} = 5^{2}$
$\Rightarrow x^{2} + y^{2} - 4 x + 6 y + 13 = 25$
$\Rightarrow x^{2} + y^{2} - 4 x + 6 y - 12 = 0,$
which is the required equation of the circle.

Suggest Corrections

Similar questions

Q. The centre of a circle is

(2, - 3)

and the circumference is

10 π

. Then, the equation of the circle is

Q. The equation of a circle with centre at

(2, - 3)

and the circumference is

10 π

units is

Q. If the lines

2 x + 3 y + 1 = 0

and

3 x - y - 4 = 0

lie along diameters of a circle of circumference

10 π

, then the equation of the circle is:

Q. lf the lines

2 x + 3 y + 1 = 0

and

3 x - y - 4 = 0

lie along diameters of a circle of circumference

10 π

, then the equation of the circle is:

Q. If the lines

2 x + 3 y + 1 = 0

and

3 x - y - 4 = 0

lie along the diameter of a circle of circumference

10 π

, then equation of circle be

Join BYJU'S Learning Program

The centre of a circle is (2, –3) and the circumference is 10π. Then the equation of the circle is

The correct option is C x2+y2−4x+6y−12=0Centre (2, – 3) Circumference = 10π ⇒ 2πr=10π ⇒ r=5. From (x−h)2+(y−k)2=r2, (x−2)2+(y+3)2=52 ⇒ x2+y2−4x+6y+13=25 ⇒ x2+y2−4x+6y−12=0, which is the required equation of the circle.

The centre of a circle is (2, –3) and the circumference is $10 π .$ Then the equation of the circle is