Circle

The general form of the equation of a circle is:

(x - h)² + (y - k)² = r²

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

Substituting the centre and radius of the circle in he general form:

⇒ (x - 2)² + (y - 4)² = 5²

⇒ (x - 2)² + (y - 4)² = 25

Ans; equation of a circle with Centre (2, 4) and radius 5 is:

⇒ (x - 2)² + (y - 4)² = 25

The general form of the equation of a circle is:

(x - h)² + (y - k)² = r²

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

Substituting the centre and radius of the circle in he general form:

⇒ (x - ( - 3))² + (y - ( - 2))² = 6²

⇒ (x + 3)² + (y + 2)² = 36

Ans; equation of a circle with Centre ( - 3, - 2) and radius 6 is:

⇒ (x + 3)² + (y + 2)² = 36

The general form of the equation of a circle is:

(x - h)² + (y - k)² = r²

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

Substituting the centre and radius of the circle in he general form:

⇒ (x - a)² + (y - a)² = (√2)²

⇒ (x - a)² + (y - a)² = 2

Ans; equation of a circle with Centre (a, a) and radius

is:

(x - a)² + (y - a)² = 2

The general form of the equation of a circle is:

(x - h)² + (y - k)² = r²

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

Substituting the centre and radius of the circle in he general form:

(x - (a cos ∝))² + (y - (a sin ∝))² = a²

⇒ (x - a cos ∝)² + (y - a sin ∝)² = a²

⇒ x² - 2xacos α + a² cos² α + y² - 2yasin α + a² sin² α = a²

⇒ x² + y² + a² (cos² α + sin² α) - 2a(xcos α + ysin α) = a²

⇒ x² + y² + a² - 2a(xcos α + ysin α) = a² …((cos²α + sin² α) = 1)

⇒ x² + y² - 2a(xcos α + ysin α) = 0

Ans: equation of a circle with Centre (a cos ∝, a sin ∝) and radius a is:

x² + y² - 2a(xcos α + ysin α) = 0

The general form of the equation of a circle is:

(x - h)² + (y - k)² = r²

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

Substituting the centre and radius of the circle in he general form:

⇒ (x - ( - a ))² + (y - ( - b))² = √(a²2 - b²2 )²

⇒ (x + a )² + (y + b)² = a² - b²

⇒ x² + 2xa + a² + y² + 2yα + b² = a² - b²

⇒ x² + 2xa + y² + 2yα = a² - 2b²

⇒ x² + y² + 2a(x + y) = a² - 2b²

⇒ x² + y² + 2a(x + y) = a² - 2b²

Ans; equation of a circle with Centre ( - a, - b) and radius

is:

⇒ x² + y² + 2a(x + y) = a² - 2b²

The general form of the equation of a circle is:

(x - h)² + (y - k)² = r²

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

Substituting the centre and radius of the circle in he general form:

⇒ (x - 0)² + (y - 0)² = 4²

⇒ x² + y² = 16

Ans; equation of a circle with . Centre at the origin and radius 4 is:

x² + y² = 16

The general form of the equation of a circle is:

(x - h)² + (y - k)² = r²

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

Comparing the given equation of circle with general form we get:

h = 3 , k = 1, r² = 9

⇒ centre = (3, 1) and radius = 3 units.

Ans: centre = (3, 1) and radius = 3 units.

The general form of the equation of a circle is:

(x - h)² + (y - k)² = r²

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

Comparing the given equation of circle with general form we get:

h = 1/2 , k = - 1/3, r² = 1/16

⇒ centre = (1/2, - 1/3) and radius = 1/4 units.

Ans: centre = (1/2, - 1/3) and radius = 1/4 units.

The general form of the equation of a circle is:

(x - h)² + (y - k)² = r²

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

Comparing the given equation of circle with general form we get:

h = - 5 , k = 3, r² = 20

⇒ centre = ( - 5, 3) and radius = √20 = 2√5 units.

Ans: centre = ( - 5, 3) and radius = 2√5 units.

The general form of the equation of a circle is:

(x - h)² + (y - k)² = r²

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

Comparing the given equation of circle with general form we get:

h = 0 , k = 1, r² = 2

⇒ centre = (0, 1) and radius = √2units.

Ans: centre = (0, 1) and radius = √2units.

The general form of the equation of a circle is:

(x - h)² + (y - k)² = r²

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

In this question we know that (h, k) = (2, - 5), so for determining the equation of the circle we need to determine the radius of the circle.

Since the circle passes through (3, 2), that pair of values for x and y must satisfy the equation and we have:

⇒ (3 - 2)² + (2 - ( - 5))² = r²

⇒ 1² + 7² = r²

⇒ r² = 49 + 1 = 50

∴ r² = 50

⇒ Equation of circle is:

(x - 2)² + (y - ( - 5))² = 50

⇒ (x - 2)² + (y + 5)² = 50

Ans:(x - 2)² + (y + 5)² = 50

The general form of the equation of a circle is:

(x - h)² + (y - k)² = r²

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

Since, centre lies on Y - axis, ∴ it’s X - coordinate = 0, i.e.h = 0

Hence, (0, k) is the centre of the circle.

Substituting the given values in general form of the equation of a circle we get,

⇒ (3 - 0)² + (2 - k)² = 5²

⇒ (3)² + (2 - k)² = 25

⇒ 9 + (2 - k)² = 25

⇒ (2 - k)² = 25 - 9 = 16

Taking square root on both sides we get,

⇒ 2 - k = ±4

⇒ 2 - k = 4 & 2 - k = - 4

⇒ k = 2 - 4 & k = 2 + 4

⇒ k = - 2 & k = 6

∴ Equation of circle when k = - 2 is:

x² + (y + 2)² = 25

Equation of circle when k = 6 is:

x² + (y - 6)² = 25

Ans: Equation of circle when k = - 2 is:

x² + (y + 2)² = 25

Equation of circle when k = 6 is:

x² + (y - 6)² = 25

The intersection of the lines: 3x + 2y = 11 and 2x + 3y = 4

Is (5, - 2)

∴ This problem is same as solving a circle equation with centre and point on the circle given.

The general form of the equation of a circle is:

(x - h)² + (y - k)² = r²

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

In this question we know that (h, k) = (2, - 3), so for determining the equation of the circle we need to determine the radius of the circle.

Since the circle passes through (5, - 2), that pair of values for x and y must satisfy the equation and we have:

⇒ (5 - 2)² + ( - 2 - ( - 3))² = r²

⇒ 3² + 1² = r²

⇒ r² = 9 + 1 = 10

∴ r² = 10

⇒ Equation of circle is:

(x - 2)² + (y - ( - 3))² = 10

⇒ (x - 2)² + (y + 3)² = 10

Ans:(x - 2)² + (y + 5)² = 10

The intersection of the lines: x – 2y = 4 and 2x + 5y + 1 = 0.

is (2, - 1)

∴ This problem is same as solving a circle equation with centre and point on the circle given.

The general form of the equation of a circle is:

(x - h)² + (y - k)² = r²

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

In this question we know that (h, k) = (2, - 1), so for determining the equation of the circle we need to determine the radius of the circle.

Since the circle passes through ( - 1, - 3), that pair of values for x and y must satisfy the equation and we have:

⇒ ( - 1 - 2)² + ( - 3 - ( - 1))² = r²

⇒ ( - 3)² + ( - 2)² = r²

⇒ r² = 9 + 4 = 13

∴ r² = 13

⇒ Equation of circle is:

(x - 2)² + (y - ( - 1))² = 13

⇒ (x - 2)² + (y + 1)² = 13

Ans:(x - 2)² + (y + 1)² = 13

The point of intersection of two diameters is the centre of the circle.

∴ point of intersection of two diameters x – y = 9 and x – 2y = 7 is (11, 2).

∴ centre = (11, 2)

Area of a circle = r²

38.5 = r²

⇒

⇒ r² = 12.25 sq.cm

the equation of the circle is:

(x - h)² + (y - k)² = r²

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

⇒ (x - 11)² + (y - 2)² = 12.25

Ans: (x - 11)² + (y - 2)² = 12.25

The equation of a circle passing through the coordinates of the end points of diameters is:

(x - x₁) (x - x₂) + (y - y₁)(y - y₂) = 0

Substituting, values:(x₁, y₁) = (3, 2) & (x₂, y₂) = (2, 5)

We get:

(x - 3)(x - 2) + (y - 2)(y - 5) = 0

⇒ x² - 2x - 3x + 6 + y² - 5y - 2y + 10 = 0

⇒ x² + y² - 5x - 7y + 16 = 0

Ans: x² + y² - 5x - 7y + 16 = 0

The equation of a circle passing through the coordinates of the end points of diameters is:

(x - x₁) (x - x₂) + (y - y₁)(y - y₂) = 0

Substituting, values:(x₁, y₁) = (5, - 3) & (x₂, y₂) = (2, - 4)

We get:

(x - 5)(x - 2) + (y + 3)(y + 4) = 0

⇒ x² - 2x - 5x + 10 + y² + 3y + 4y + 12 = 0

⇒ x² + y² - 7x + 7y + 22 = 0

Ans: x² + y² - 7x + 7y + 22 = 0

The equation of a circle passing through the coordinates of the end points of diameters is:

(x - x₁) (x - x₂) + (y - y₁)(y - y₂) = 0

Substituting, values:(x₁, y₁) = ( - 2, - 3) & (x₂, y₂) = ( - 3, 5)

We get:

(x + 2)(x + 3) + (y + 3)(y - 5) = 0

⇒ x² + 3x + 2x + 6 + y² - 5y + 3y - 15 = 0

⇒ x² + y² + 5x - 2y - 9 = 0

Ans: x² + y² + 5x - 2y - 9 = 0

The equation of a circle passing through the coordinates of the end points of diameters is:

(x - x₁) (x - x₂) + (y - y₁)(y - y₂) = 0

Substituting, values:(x₁, y₁) = (p, q) & (x₂, y₂) = (r, s)

We get:

(x - p)(x - r) + (y - q)(y - s) = 0

⇒ x² - rx - px + pr + y² - sy - qy + qs = 0

⇒ x² + y² - (r + p)x - (s + q)y + (pr + qs) = 0

Ans: x² + y² - (r + p)x - (s + q)y + (pr + qs) = 0

The intersection points in clockwise fashion are:( - 2, 5), (4, 5), (4, - 2), ( - 2, - 2).

The equation of a circle passing through the coordinates of the end points of diameters is:

(x - x₁) (x - x₂) + (y - y₁)(y - y₂) = 0

Substituting, values:(x₁, y₁) = ( - 2, 5) & (x₂, y₂) = (4, - 2)

We get:

(x + 2)(x - 4) + (y - 5)(y + 2) = 0

⇒ x² - 4x + 2x - 8 + y² + 2y - 5y - 10 = 0

⇒ x² + y² - 2x - 3y - 18 = 0

Ans:x² + y² - 2x - 3y - 18 = 0

The general equation of a conic is as follows

ax² + 2hxy + by² + 2gx + 2fy + c = 0 where a, b, c, f, g, h are constants

For a circle, a = b and h = 0.

The equation becomes:

x² + y² + 2gx + 2fy + c = 0…(i)

Given, x² + y² – 4x + 6y – 5 = 0

Comparing with (i) we see that the equation represents a circle with 2g = - 4 g = - 2, 2f = 6f = 3 and c = - 5.

Centre ( - g, - f) = { - ( - 2), - 3}

= (2, - 3).

Radius =

The general equation of a conic is as follows

ax² + 2hxy + by² + 2gx + 2fy + c = 0 where a, b, c, f, g, h are constants

For a circle, a = b and h = 0.

The equation becomes:

x² + y² + 2gx + 2fy + c = 0…(i)

Given, x² + y² + x - y = 0

Comparing with (i) we see that the equation represents a circle with 2g = 1 , 2f = - 1 and c = 0.

Centre ( - g, - f) =

= ().

Radius =

The general equation of a conic is as follows

ax² + 2hxy + by² + 2gx + 2fy + c = 0 where a, b, c, f, g, h are constants

For a circle, a = b and h = 0.

The equation becomes:

x² + y² + 2gx + 2fy + c = 0…(i)

Given, 3x² + 3y² + 6x - 4y – 1 = 0

Comparing with (i) we see that the equation represents a circle with 2g = 2 g = 1, 2f = and .

Centre ( - g, - f) = { - 1, - ()}

= ( - 1, ).

Radius =

The general equation of a circle:

x² + y² + 2gx + 2fy + c = 0…(i) where c, g, f are constants.

Given, x² + y² + 2x + 10y + 26 = 0

Comparing with (i) we see that the equation represents a circle with 2g = 2g = 1, 2f = 10f = 5 and c = 26.

Centre ( - g, - f) = ( - 1, - 5).

Radius =

Thus it is a point circle with radius 0.

Radius =

which implies that the radius is negative. (not possible)

Therefore, x² + y² - 3x + 3y + 10 = 0 does not represent a circle.

(i) The required circle equation

= 0

Using Laplace Expansion

is the equation with centre = (2.5, 0.5)

Radius =

(ii) The required circle equation

= 0

Using Laplace Expansion

is the equation with centre = (11, 2)

Radius =

(iii) The required circle equation

= 0

Using Laplace Expansion

is the equation with centre = (7, 3)

Radius =

The general equation of a circle: (x - h)² + (y - k)² = r²

…(i), where (h, k) is the centre and r is the radius.

Putting A( - 2, 3), B(5, 2) and c(6, - 1) in (i) we get

h² + k² + 4h - 6k + 13 = r² …(ii)

h² + k² - 10h - 4k + 29 = r² …(iii)and

h² + k² - 12h + 2k + 37 = r² …(iv)

subtracting (ii) from (iii) and also from (iv),

- 14h + 2k + 16 = 0 - 7h + k + 8 = 0

- 16h + 8k + 24 = 0 - 2h + k + 3 = 0

Subtracting,

5h - 5 = 0h = 1

k = - 1

Centre = (1, - 1)

Putting these values in (ii) we get, radius = = 5

Equation of the circle is

(x - 1)² + {y - ( - 1)}² = 5²

(x - 1)² + (y + 1)² = 25.

2 or more circles are said to be concentric if they have the same centre and different radii.

Given, x² + y² + 4x + 6y + 11 = 0

The concentric circle will have the equation

x² + y² + 4x + 6y + c’ = 0

As it passes through P(5, 4), putting this in the equation

5² + 4² + 45 + 64 + c’ = 0

25 + 16 + 20 + 24 + c’ = 0

c’ = - 85

The required equation is

x² + y² + 4x + 6y - 85 = 0

The general equation of a circle: (x - h)² + (y - k)² = r²

…(i), where (h, k) is the centre and r is the radius.

Putting (1, 0) in (i)

(1 - h)² + (0 - k)² = r²

h² + k² + 1 - 2h = r² ..(ii)

Putting (2, - 7) in (i)

(2 - h)² + ( - 7 - k)² = r²

h² + k² + 53 - 4h + 14k = r²

(h² + k² + 1 - 2h) + 52 - 2h + 14k = r²

h - 7k - 26 = 0..(iii) [from (ii)]

Similarly putting (8, 1)

7h + k - 32 = 0..(iv)

Solving (iii)&(iv)

h = 5 and k = - 3

centre(5, - 3)

Radius = 25

To check if (9, - 6) lies on the circle, (9 - 5)² + ( - 6 + 3)² = 5²

Hence, proved.

The equation of a circle:

x² + y² + 2gx + 2fy + c = 0…(i)

Putting (1, 3) & (2, - 1)in (i)

2g + 6f + c = - 10..(ii)

4g - 2f + c = - 5..(iii)

Since the centre lies on the given straight line, ( - g, - f) must satisfy the equation as

- 2g –f - 4 = 0…(iv)

Solving, f = - 1, g = - 1.5, c = - 1

The equation is

x² + y² - 3x - 2y - 1 = 0

The given image of the circle is:

We know that the general equation of the circle is given by:

x² + y² + 2gx + 2fy + c = 0

Also,

Radius r =

Now,

r = 4 units.

We need to the find the equation of the circle which is concentric to the given circle and touches y-axis.

The centre of the circle remains the same.

Now, y-axis will be tangent to the circle.

Point of contact will be (0, 3)

Therefore, radius = 2

Now,

Equation of the circle:

(x – 2)² + (y – 3)² = (2)²

x² + 4 – 4x + y² + 9 – 6y = 4

x² + y² – 4x – 6y + 9 = 0

2 or more circles are said to be concentric if they have the same centre and different radii.

Given, x² + y² - 6x + 12y + 15 = 0

Radius r =

The concentric circle will have the equation

x² + y² - 6x + 12y + c’ = 0

Also given area of circle = 2area of the given circle.

r’² = 2r² = 230 = 60

We can get c’ = 45 - 60 = - 15

The required equation is x² + y² - 6x + 12y - 15 = 0.

Given,

x² + y² – 4x – 6y – 12 = 0

centre ( - g₁, - f₁) = (2, 3)

x² + y² + 2x + 4y – 5 = 0

centre ( - g₂, - f₂) = ( - 1, - 2)

x² + y² – 10x – 16y + 7 = 0

centre ( - g₃, - f₃) = (5, 8)

to prove that the centres are collinear,

Where x₁, y₁ are the coordinates of the ist centre and so on.

= 2( - 2 - 8) - 3( - 1 - 5) + 1( - 8 + 10)

= - 20 + 18 + 2 = 0

The centres are collinear.

The general equation of a circle: (x - h)² + (y - k)² = r²

…(i), where (h, k) is the centre and r is the radius.

Putting A(1, 1) in (i)

(1 - h)² + (1 - k)² = 1²

h² + k² + 2 - 2h - 2k = 1

h² + k² - 2h - 2k = - 1..(ii)

Putting B(2, 2) in (i)

(2 - h)² + (2 - k)² = 1²

h² + k² + 8 - 4h - 4k = 1

h² + k² - 4h - 4k = - 7

(h² + k² - 2h - 2k) - 2h - 2k = - 7

- 1 - 2h - 2k = - 7 [from (ii)]

- 2h - 2k = - 6

h + k = 3 h= 3 - k

Putting it in (ii)

(3 - k)² + k² - 2(3 - k) - 2k = - 1

9 + 2k² - 6k - 6 + 2k - 2k = - 1

2k² + 4 - 6k = 0

k² - 3k + 2 = 0

k = 2 or k = 1

When k = 2, h = 3 - 2 = 1

Equation of 1 circle

(x - 1)² + (y - 2)² = 1

When k = 1, h = 3 - 1 = 2

(x - 2)² + (y - 1)² = 1

From the figure

AD = b units and AE = a units.

D(0, b), E(a, 0) and A(0, 0) lies on the circle. C is the centre.

The general equation of a circle: (x - h)² + (y - k)² = r²

…(i), where (h, k) is the centre and r is the radius.

Putting A(0, 0) in (i)

(0 - h)² + (0 - k)² = r²

h² + k² = r² …(ii)

Similarly putting D(0, b) in (i)

(0 - h)² + (b - k)² = r²

h² + k² + b² - 2kb = r²

r² + b² - 2kb = r²

b² - 2kb = 0

(b- 2k)b = 0

Either b = 0ork =

Similarly putting E(a, 0) in (i)

(a - h)² + (0 - k)² = r²

h² + k² + a² - 2ha = r²

r² + a² - 2ha = r²

a² - 2ha = 0

(a- 2h)a = 0

Either a = 0orh =

Centre = C

r² = h² + k²

Putting the value of r² , h and k in equation (i)

(x - h)² + (y - k)² = r²

which is the required equation.

Solving the equations we get the coordinates of the triangle:

The required circle equation

= 0

Using Laplace Expansion

.

Solving the euations we get the coordinates of the quadrilateral.

Slope of = 1

Slope of CD = 1

AB||CD

Slope of BD = AC = - 1

AC||BD

So they form a rectangle with all sides = 90

The quadrilateral is cyclic as sum of opposite angles = 180.

Now, AD = diameter of the circle equation of the circle with extremities A(0, 0)&D(6, - 4) is

(x - 0)(x - 6) + (y - 0)(y + 4) = 0

x² + y² – 6x + 4y = 0

Given x² + y² – 6x + 5y – 7 = 0

Centre( 3, )

As ( - 1, 3) & (∝, β) are the 2 extremities of the diameter, using mid - point formula we can write

and

(∝, β) = (7, - 8)