- Find the equation of the circle with centre (0, 2) and radius 2
Solution:
The equation of a circle with centre (h, k) and radius r is given as (x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (0, 2) and radius (r) = 2.
Therefore, the equation of the circle is
(x – 0)2 + (y – 2)2 = 22
⟹ x2 + y2 + 4 – 4 y = 4
⟹ x2 + y2 – 4y = 0
- Find the equation of the circle with centre (–2, 3) and radius 4
Solution:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (–2, 3) and radius (r) = 4.
Therefore, the equation of the circle is
(x + 2)2 + (y – 3)2 = (4)2
⟹ x2 + 4x + 4 + y2 – 6y + 9 = 16
⟹ x2 + y2 + 4x – 6y – 3 = 0
- Find the equation of the circle with centre(1/2, 1/4) and radius(1/12)
Solution:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) =(1/2, 1/4) and radius (r) = (1/12)
Therefore, the equation of the circle is
(x – 1/2)2 + (y – 1/4)2 = (1/12)2
x2 – x + 1/4 + y2 – y/2 + 1/16 = 1/144
x2 – x + 1/4 + y2 – y/2 + 1/16 – 1/144 = 0
144x2 – 144x + 36 + 144y2 – 72y + 44 = 0
36x2 – 36x + 36y2 – 18y + 11 = 0
36x2 + 36y2 – 36x – 18y + 11=0
- Find the equation of the circle with centre (1, 1) and radius √2.
Solution:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (1, 1) and radius (r) = √2.
Therefore, the equation of the circle is
(x – 1)2 + (y – 1)2 = (√2)2
x2 – 2x + 1+y2 – 2y + 1 = 2
x2 + y2 -2x – 2y =0
- Find the equation of the circle with centre (-a, -b) and radius √(a2 – b2).
Solution:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (-a, -b) and radius (r) = √(a2 – b2).
Therefore, the equation of the circle is
(x + a)2 + (y + b)2 = (√(a2 – b2))2
x2 + 2ax + a2 +y2 + 2by + b2 = a2 – b2
x2 + y2 + 2ax + 2by + 2b2 = 0
- Find the centre and radius of the circle (x + 5)2 + (y – 3)2 = 36
Solution:
The equation of the given circle is (x + 5)2 + (y – 3)2 = 36.
(x + 5)2 + (y – 3)2 = 36
⇒ {x – (–5)}2 + (y – 3)2 = 62, which is of the form (x – h)2 + (y – k)2 = r2, where h = – 5, k = 3, and r = 6.
Thus, the centre of the given circle is (–5, 3), while its radius is 6.
- Find the centre and radius of the circle x2 + y2 – 4x – 8y – 45 = 0
Solution:
The equation of the given circle is x2 + y2 – 4x – 8y – 45 = 0.
x2 + y2 – 4x – 8y – 45 = 0
⇒ (x2 – 4x) + (y2 – 8y) = 45
⇒ {x2 – 2(x)(2) + 22} + {y2 – 2(y)(4)+ 42} – 4 –16 = 45
⇒ (x – 2)2 + (y –4)2 = 65
⇒ (x – 2)2 + (y – 4)2= 65 ,
Which is of the form (x – h)2 + (y – k)2 = r2, where h = 2, k = 4 and since r2 = 65, we have r = √65 ,
Thus, the centre of the given circle is (2, 4), while its radius is √65 .
- Find the centre and radius of the circle x2 + y2 – 8x + 10y – 12 = 0
Solution:
The equation of the given circle is x2 + y2 – 8x + 10y – 12 = 0.
x2 + y2 – 8x + 10y – 12 = 0
⇒ (x2 – 8x) + (y2 + 10y) = 12
⇒ {x2 – 2(x)(4) + 42} + {y2 + 2(y)(5) + 52}– 16 – 25 = 12
⇒ (x – 4)2 + (y + 5)2 = 53
⟹ (x − 4)2 +{y − (−5)}2 = (√53)2 which is of the form (x – h)2 + (y – k)2 = r2, where h = 4, k = – 5 and r = √53
Thus, the centre of the given circle is (4, –5), while its radius is √53.
- Find the centre and radius of the circle 2x2 + 2y2 – x = 0
Solution:
The equation of the given circle is 2x2 + 2y2 – x = 0.
2x2 + 2y2 – x = 0
(2x2 – x)+2y2 = 0
2[(x2 – x/2)+y2] = 0
{x2 – 2x(1/4)+(1/4)2 } + y2 – (1/4)2 = 0
(x – 1/4)2 + (y – 0)2 = (1/4)2
which is of the form (x – h)2 + (y – k)2 = r2, where h = 1/4, k = 0 and r = 1/4.
Thus, the centre of the given circle is (1/4, 0), while its radius is 1/4.
- Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Solution:
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the circle passes through points (4, 1) and (6, 5),
(4 – h)2 + (1 – k)2 = r2 …………………. (1)
(6 – h)2 + (5 – k)2 = r2 …………………. (2)
Since the centre (h, k) of the circle lies on line 4x + y = 16,
4h + k = 16 ———————(3)
From equations (1) and (2), we obtain
(4 – h)2 + (1 – k)2 = (6 – h)2 + (5 – k)2
⇒ 16 – 8h + h2 + 1 – 2k + k2 = 36 – 12h + h2 + 25 – 10k + k2
⇒ 16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k
⇒ 4h + 8k = 44
⇒ h + 2k = 11 ———————–(4)
On solving equations (3) and (4), we obtain h = 3 and k = 4.
On substituting the values of h and k in equation (1), we obtain (4 – 3)2 + (1 – 4)2 = r2
⇒ (1)2 + (– 3)2 = r2
⇒ 1 + 9 = r2
⇒ r2 = 10
⇒ r = √10
Thus, the equation of the required circle is
(x – 3)2 + ( y – 4)2 = (√10)2
x2 – 6x + 9 + y2 – 8y + 16 = 10
x2 – 6x + y2 – 8y + 15 = 0
- Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.
Solution:
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the circle passes through points (2, 3) and (–1, 1),
(2 – h)2 + (3 – k)2 = r2——————- (1)
(–1 – h)2 + (1 – k)2 = r2 ————–(2)
Since the centre (h, k) of the circle lies on line x – 3y – 11 = 0,
h – 3k = 11 ———–(3)
From equations (1) and (2), we obtain
(2 – h)2 + (3 – k)2 = (–1 – h)2 + (1 – k)2
⇒ 4 – 4h + h2 + 9 – 6k + k2 = 1 + 2h + h2 + 1 – 2k + k2
⇒ 4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k
⇒ 6h + 4k = 11 ————(4)
On solving equations (3) and (4), we obtain h =7/2 and k =−5/2
On substituting the values of h and k in equation (1), we get
(2 – 7/2)2 + (3 + 5/2)2 = r2
(4-7/2)2 + (6+5/2)2 = r2
(-3/2)2 + (11/2)2 = r2
9/4+ 121/4 = r2
130/4 = r2
Thus the equation of the required circle is
(x – 7/2)2 + (y + 5/2)2 = 130/4
(2x-7/2)2 + (2y+5/2)2 = 130/4
4x2 – 28x + 49 + 4y2 + 20y + 25 = 130
4x2 + 4y2 – 28x + 20y – 56 = 0
4(x2 + y2 – 7x + 5y – 14) = 0
x2 + y2 – 7x + 5y – 14 = 0
- Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Solution:
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.
Now, the equation of the circle becomes (x – h)2 + y2 = 25.
It is given that the circle passes through point (2, 3).
( 2 – h)2 + 32 = 25
( 2 – h)2 = 25 – 9 = 16
2 – h = ±√16 = ±4
If 2 – h = 4 then h = -2
If 2 – h = -4 then h = 6
when h = -2 the equation of the circle becomes
(x + 2)2 + y2 = 25
x2 + 4x + 4 + y2 – 25 = 0
x2 + 4x + y2 – 21 = 0
When h = 6 the equation of the circle becomes
(x – 6)2 + y2 = 25
x2 – 12x + 36 + y2 – 25 = 0
x2 – 12x + y2 + 11 = 0
- Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.
solution:
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the centre of the circle passes through (0, 0),
(0 – h)2 + (0 – k)2 = r2
⇒ h2 + k2 = r2
The equation of the circle now becomes (x – h)2 + ( y- k)2 =h2 + k2
It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,
(a – h)2 + (0 – k)2 = h2 + k2————- (1)
(0 – h)2 + (b – k)2 = h2 + k2————— (2)
From equation (1), we obtain a2 – 2ah + h2 + k2 = h2 + k2
⇒ a2 – 2ah = 0
⇒ a(a – 2h) = 0
⇒ a = 0 or (a – 2h) = 0
However, a ≠ 0; hence, (a – 2h) = 0 ⇒ h =a/2.
From equation (2), we obtain h2 + b2 – 2bk + k2 = h2 + k2
⇒ b2 – 2bk = 0
⇒ b(b – 2k) = 0
⇒ b = 0 or(b – 2k) = 0
However, b ≠ 0; hence, (b – 2k) = 0 ⇒ k =b/2.
Thus, the equation of the required circle is
(x – a/2)2 + (y – b/2 )2 = ( a/2)2 +( b/2)2
(2x-a/2)2 + (2y-b/2 )2 = [a2 +b2]/2
4x2 – 4ax +a2 + 4y2 – 4by + b2 = a2 + b2
4x2+ 4y2– 4ax– 4by = 0
x2+ y2– ax – by = 0
- Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).
Solution:
The centre of the circle is given as (h, k) = (2, 2).
Since the circle passes through point (4, 5), the radius (r) of the circle is the distance between the points (2, 2) and (4, 5).
r = √[(2-4)2 + (2-5)2] = √[(2)2 + (3)2] = √(4+9) = √13
Thhus the equation of the circle is
( x- h)2 + ( y-k)2 = r2
( x- 2)2 + ( y-2)2 = (√13)2
x2 – 4x + 4 +y2 -4y + 4 = 13
x2 + y2– 4x-4y-5 = 0
- Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?
Solution:
The equation of the given circle is x2 + y2 = 25.
x2 + y2 = 25
⇒ (x – 0)2 + (y – 0)2 = 52,which is of the form (x – h)2 + (y – k)2 = r2, where h = 0, k = 0, and r = 5.
∴ Centre = (0, 0) and radius = 5
Distance between point (–2.5, 3.5) and centre (0, 0)
=√[(-2.5-0)2+(3.5-0)2]
=√(6.25+12.25)
=√18.5
≈4.3 <5
Since the distance between point (–2.5, 3.5) and centre (0, 0) of the circle is less than the radius of the circle, point (–2.5, 3.5) lies inside the circle.
