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
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