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 xaxis 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 coordinates 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
11 and 12 Grade Math
From Problems on Circle to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.