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प्रश्न
The equation of a circle with radius 5 and touching both the coordinate axes is
पर्याय
x2 + y2 ± 10x ± 10y + 5 = 0
x2 + y2 ± 10x ± 10y = 0
x2 + y2 ± 10x ± 10y + 25 = 0
x2 + y2 ± 10x ± 10y + 51 = 0
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उत्तर
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.
The equation of a circle that touches both the coordinate axes and has radius a is
The equation of a circle that touches both the coordinate axes and has radius a is \[x^2 + y^2 - 2ax + 2ay + a^2 = 0\].
Hence, the required equation of the circle is x2 + y2 ± 10x ± 10y + 25 = 0.
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