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Question
Find the equation circle if the equations of two diameters are 2x + y = 6 and 3x + 2y = 4. When radius of circle is 9
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Solution
Given equations of diameters are 2x + y = 6 and 3x + 2y = 4.
Let C (h, k) be the centre of the required circle. Since point of intersection of diameters is the centre of the circle,
x = h, y = k
∴ Equations of diameters become
2h + k = 6 …(i)
and 3h + 2k = 4 …(ii)
By (ii) – 2 x (i), we get
– h = – 8
∴ h = 8
Substituting h = 8 in (i), we get
2(8) + k = 6
∴ k = 6 – 16
∴ k = – 10
∴ Centre of the circle is C (8, –10) and radius, r = 9
The equation of a circle with centre at (h, k) and radius r is given by
(x – h)2 + (y – k)2 = r2
Here, h = 8, k = –10
∴ The required equation of the circle is
(x – 8)2 + (y +10)2 = 92
∴ x2 – 16x + 64 + y2 + 20y + 100 = 81
∴ x2 + y2 – 16x + 20y + 100 + 64 – 81 = 0
∴ x2 + y2 – 16x + 20y + 83 = 0.
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