Problems on Circle

We will learn how to solve different types of problems on circle.

1. Find the equation of a circle of radius 5 whose centre lies on x-axis and passes through the point (2, 3).

Solution:

Let the coordinates of the centre of the required circle be C(a, 0). Since it passes through the point P(2, 3).

Therefore, CP = radius

⇒ CP = 5

⇒ $\mathrm{\sqrt{(a - 2)^{2} + (0 - 3)^{2}}}$ = 5

⇒ (a - 2)$^{2}$ + 9 = 25

⇒ (a - 2)$^{2}$ = 25 - 9

⇒ (a - 2)$^{2}$ = 16

⇒ a - 2 = ± 4

⇒ a = -2 or 6

Thus, the coordinates of the centre are (-2, 0) and (6, 0).

Hence, the equation of the required circle are

(x - 2)$^{2}$ + (y – 0)^2 = 5^2 and (x – 6)$^{2}$ + (y – 0)$^{2}$ = 5$^{2}$

⇒ x$^{2}$ + y$^{2}$ + 4x – 21 = 0 and x$^{2}$ + y$^{2}$ – 12x + 11 = 0

2. Find the equation of the circle which passes through the points (3, 4) and (- 1, 2) and whose centre lies on the line x - y = 4.

Solution:       

Let the equation of the required circle be

x$^{2}$ + y$^{2}$ + 2gx + 2fy + c = 0 ............... (i)

According to the problem the equation (i) passes through the points (3, 4) and (- 1, 2). Therefore,

9 + 16 + 6g + 8f + c = 0 ⇒ 6g + 8f + c = - 25 ............... (ii)

and 1 + 4 - 2g + 4f + c = 0 ⇒ - 2g + 4f + c = - 5 ............... (iii)

Again according to the problem, the centre of the circle (i) lies on the line x - y = 4.

Therefore,

- g  - (- f) = 4               

⇒ - g + f = 4 ............... (iv)

Now, subtract the equation (iii) from (ii) we get,    

8g + 4f = - 20     

⇒ 2g + f = - 5 ............... (v)

Solving equations (iv) and (v) we get, g = - 3 and f = 1.

Putting g = - 3 and f = 1 in (iii) we get, c = -15.

Therefore, the equation of the requited circle is x$^{2}$ + y$^{2}$ - 6x + 2y - 15 = 0.

More problems on circle:

3. Find the equation to the circle described on the common chord of the given circles x$^{2}$ + y$^{2}$ - 4x - 5 = 0 and x$^{2}$ + y$^{2}$ + 8x + 7 = 0 as diameter. 

Solution:           

Let, S$_{1}$ = x$^{2}$ + y$^{2}$ - 4x - 5 = 0 ............... (i)

and S$_{2}$ = x$^{2}$ + y$^{2}$ + 8x + 7 = 0 ............... (ii)

Then, the equation of the common chord of the circles (1) and (2) is,

S$_{2}$ - S$_{1}$ = 0

⇒ 12x + 12 = 0    

⇒ x + 1 = 0 ............... (iii)

Let the equation of the circle described on the common chord of (i) and (ii) as diameter be

x$^{2}$ + y$^{2}$  - 4x - 5 + k(x + 1) = 0

⇒ x$^{2}$ + y$^{2}$  - (4 - k)x - 5 + k = 0 ............... (iv)

Clearly, the co-ordinates of the centre of the circle (4) are ($\frac{4 - k}{2}$, 0) Since the common chord (iii) is a diameter of the circle (iv) hence,

$\frac{4 - k}{2}$ + 1 = 0     

⇒ k = 6.  

Now putting the value of k = 6 in x$^{2}$ + y$^{2}$ - (4 - k) x- 5 + k = 0 we get,

x$^{2}$ + y$^{2}$  - (4 - 6) x - 5 + 6 = 0

⇒ x$^{2}$ + y$^{2}$ + 2x + 1 = 0, which is the required equation of the circle.

● The Circle

• Definition of Circle
• Equation of a Circle
• General Form of the Equation of a Circle
• General Equation of Second Degree Represents a Circle
• Centre of the Circle Coincides with the Origin
• Circle Passes through the Origin
• Circle Touches x-axis
• Circle Touches y-axis
• Circle Touches both x-axis and y-axis
• Centre of the Circle on x-axis
• Centre of the Circle on y-axis
• Circle Passes through the Origin and Centre Lies on x-axis
• Circle Passes through the Origin and Centre Lies on y-axis
• Equation of a Circle when Line Segment Joining Two Given Points is a Diameter
• Equations of Concentric Circles
  
• Circle Passing Through Three Given Points
• Circle Through the Intersection of Two Circles
• Equation of the Common Chord of Two Circles
• Position of a Point with Respect to a Circle
• Intercepts on the Axes made by a Circle
• Circle Formulae
• Problems on Circle

11 and 12 Grade Math 

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