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Find the Equation of the Circle Whose Diameter is the Line Segment Joining (−4, 3) and (12, −1). Find Also the Intercept Made by It on Y-axis.

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प्रश्न

Find the equation of the circle whose diameter is the line segment joining (−4, 3) and (12, −1). Find also the intercept made by it on y-axis.

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उत्तर

It is given that the end points of the diameter of the circle are (−4, 3) and (12, −1).
∴ Required equation of circle: \[\left( x + 4 \right)\left( x - 12 \right) + \left( y - 3 \right)\left( y + 1 \right)\]

or

\[x^2 + y^2 - 8x - 2y - 51 = 0\]  ...(1)
Putting x = 0 in (1):
y2 − 2y − 51 = 0
⇒ y2 −  2y − 51 = 0
⇒ \[y = 1 \pm 2\sqrt{13}\]
Hence, the intercepts made by it on the y-axis is
\[1 + 2\sqrt{13} - 1 + 2\sqrt{13} = 4\sqrt{13}\]
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Advanced Concept of Circle - Standard Equation of a Circle
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अध्याय 24: The circle - Exercise 24.3 [पृष्ठ ३७]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 24 The circle
Exercise 24.3 | Q 7 | पृष्ठ ३७

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