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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. [Ncert Exemplar] - Mathematics

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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.

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 − 10+ 25 = 0

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Circle - Standard Equation of a Circle
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अध्याय 24: The circle - Exercise 24.2 [पृष्ठ ३२]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 24 The circle
Exercise 24.2 | Q 15 | पृष्ठ ३२

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