Maths NCERT Solutions Class 11 | CBSE Class 11th Mathematics
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Class 11 maths ncert solutions – Circle
Circle: A circle is the set of all points in a plane which is at a fixed distance from a fixed point in the plane. The fixed point is called the centre of the circleand the distance from the centre to any point on the circle is called the radius of the circle.
In this figure,C is the centre and CP is a radius.
- The equation of a circle with radius r having centre (h, k) is given by
(x – h)2 + (y – k)2= r2
- The general equation of the circle is given by x2+ y2 + 2gx + 2fy + c = 0, where g, f and c
(a) The centre of this circle is (– g, – f)
(b) The radius of the circle is
Question: Find the equation of the circle with centre (1, 2) and radius 3.
Solution: We have h = 1 ,k = 2 and r = 3
So equation of the circle is (x-1)2 + (y-2)2 = 32
or(x-1)2 + (y-2)2=9
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Question: Find the centre and radius of the circle x2 + y2 -2x+ 4y – 8 = 0.
Solution: First of all,we write the given equation in the form (x2 – 2x) + ( y2 + 4y) = 8
Now, completing the squares, we get
(x2 – 2x + 1) + ( y2 + 4y + 4) = 8 + 1 + 4
(x – 1)2+ (y + 2)2 = 13
Comparing it with the standard form of the equation of the circle, we see that the
centre of the circle is (1, –2) and radius is .
Comparing it with the general equation of the circle , we have
2g = -2 or g = -1
2f = 4 or f = 2
c = -8
So centre of circle is (1,-2) and
Radius is = .
Question: Find the equation of a circle passing through(4,1) and (6,5) and whose centre is on the line 4x+ y = 16.
Solution: Let the equation of circle is x2+ y2 + 2gx + 2fy + c = 0
Since it passes through (4,1) so it satisfies the equation of circle
ie 16+1+8g+2f+c = 0
or 8g+2f+c = -17 _ (1)
Also it passes through (6,5) so we get
12g+10f+c = -61 _(2)
Now centre is on line 4x+ y = 16 and we know centre is (-g,-f)
So it must satisfies the equation of line
ie -4g-f = 16 _(3)
Solving the above three equations,we get g =-3, f = – 4 and c = 15
Hence the equation of circle is
x2+ y2-6x-8y+15 = 0