मराठी

Find the Equation of the Circle Circumscribing the Rectangle Whose Sides Are X − 3y = 4, 3x + Y = 22, X − 3y = 14 and 3x + Y = 62.

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प्रश्न

Find the equation of the circle circumscribing the rectangle whose sides are x − 3y = 4, 3x + y = 22, x − 3y = 14 and 3x + y = 62.

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उत्तर

Given:
Sides of the rectangle:
x − 3y = 4     ...(1)
3x + y = 22   ...(2)
x − 3y = 14    ...(3)
And, 3x + y = 62    ...(4)

The intersection of (1) and (2) is (7, 1).
The intersection of (2) and (3) is (8, −2).
The intersection of (3) and (4) is (20, 2).
The intersection of (1) and (4) is (19, 5).
Hence, the vertices of the rectangle are (7, −1), (8, −2), (20, 2) and (19, 5).
The vertices of the diagonals are (7, −1), (20, 2) and (19, 5), (8, −2).
Thus, the required equation of the circle is

\[\left( x - 7 \right)\left( x - 20 \right) + \left( y - 1 \right)\left( y - 2 \right) = 0\] or
\[x^2 + y^2 - 27x - 3y + 142 = 0\]
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Advanced Concept of Circle - Standard Equation of a Circle
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पाठ 24: The circle - Exercise 24.3 [पृष्ठ ३७]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 24 The circle
Exercise 24.3 | Q 4 | पृष्ठ ३७

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