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Conic Sections — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsConic Sections1 Mark
MCQ
The equation of a circle with radius $5$ and touching both the coordinate axes is:
  • A
    $x^2+ y^2± 10x ± 10y + 5 = 0$
  • B
    $x^2+ y^2 ± 10x ± 10y = 0$
  • ✓
    $x^2+ y^2± 10x ± 10y + 25 = 0$
  • D
    $x^2+ y^2 ± 10x ± 10y + 51 = 0$
✓

Answer

Correct option: C.
$x^2+ y^2± 10x ± 10y + 25 = 0$
Case $I:$ If the circle lies in the first quadrant:
The equation of a circle that touches both the coordinate axes and hasradius 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 $x^2+ y^2± 10x ± 10y + 25 = 0.$

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