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प्रश्न
Find the equation of the circle which passes through the points (2, 3) and (4,5) and the centre lies on the straight line y − 4x + 3 = 0.
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उत्तर
The general equation of the circle is x2 + y2 + 2gx + 2fy + c = 0 where the centre of the circle is (−g, −f)
Now, it is passing through (2, 3)
∴ 13 + 4g + 6f + c = 0 .....(1)
Also, it is passing through (4, 5)
∴ 41 + 8g + 10f + c = 0 .....(2)
⇒g=−a2
Now, the centre lies on the straight line y − 4x + 3 = 0
∴ −f + 4g + 3 = 0 .....(3)
⇒g=−a2
Solving (1), (2) and (3), we get
g = −2, f = −5 and c = 25
The equation of the circle is given by x2 + y2 − 4x − 10y + 25 = 0
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