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The Equation of a Circle with Radius 5 and Touching Both the Coordinate Axes is - Mathematics

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Question

The equation of a circle with radius 5 and touching both the coordinate axes is

Options

  • x2 + y2 ± 10x ± 10y + 5 = 0

  • x2 + y2 ± 10x ± 10y = 0

  • x2 + y2 ± 10x ± 10y + 25 = 0

  • x2 + y2 ± 10x ± 10y + 51 = 0

MCQ
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Solution

x2 + y2 ± 10x ± 10y + 25 = 0

Case I: If the circle lies in the first quadrant:
The equation of a circle that touches both the coordinate axes and has radius a is \[x^2 + y^2 - 2ax - 2ay + a^2 = 0\].

The given radius of the circle is 5 units, i.e.

\[a = 5\].

 Thus, the equation of the circle is \[x^2 + y^2 - 10x - 10y + 25 = 0\].

Case II: If the circle lies in the second quadrant:
The equation of a circle that touches both the coordinate axes and has radius a is \[x^2 + y^2 + 2ax - 2ay + a^2 = 0\].

The given radius of the circle is 5 units, i.e.

\[a = 5\].
  Thus, the equation of the circle is
\[x^2 + y^2 + 10x - 10y + 25 = 0\]
Case III: If the circle lies in the third quadrant:
The equation of a circle that touches both the coordinate axes and has radius a is
\[x^2 + y^2 + 2ax + 2ay + a^2 = 0\] .
The given radius of the circle is 5 units, i.e.
\[a = 5\].
 Thus, the equation of the circle is \[x^2 + y^2 + 10x + 10y + 25 = 0\].
Case IV: If the circle lies in the fourth quadrant:
The equation of a circle that touches both the coordinate axes and has radius a is \[x^2 + y^2 - 2ax + 2ay + a^2 = 0\].
The given radius of the circle is 5 units, i.e.
\[a = 5\].
 Thus, the equation of the circle is \[x^2 + y^2 - 10x + 10y + 25 = 0\].

Hence, the required equation of the circle is x2 + y2 ± 10x ± 10y + 25 = 0.

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Advanced Concept of Circle - Standard Equation of a Circle
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Chapter 24: The circle - Exercise 24.6 [Page 40]

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RD Sharma Mathematics [English] Class 11
Chapter 24 The circle
Exercise 24.6 | Q 14 | Page 40

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