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CONIC SECTIONS — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCONIC SECTIONS1 Mark
MCQ
The equation of a circle with radius $5$ and touching both the coordinate axes is:
  • A
    $x^2+y^2 \pm 10 x \pm 10 y+5=0$
  • B
    $x^2+y^2 \pm 10 x \pm 10 y=0$
  • ✓
    $x^2+y^2 \pm 10 x \pm 10 y+25=0$
  • D
    $x^2+y^2 \pm 10 x \pm 10 y+51=0$
✓

Answer

Correct option: C.
$x^2+y^2 \pm 10 x \pm 10 y+25=0$
  1. $x^2+y^2 \pm 10 x \pm 10 y+25=0$
Solution:
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 $x^2+y^2-2 a x-2 a y+a^2=0$.
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is $x^2+y^2-10 x-10 y+25=0$.
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 $x^2+y^2+2 a x-2 a y+a^2=0$.
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is $x^2+y^2+10 x-10 y+25=0$.
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 $x^2+y^2+2 a x+2 a y+a^2=0$
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is $x^2+y^2+10 x+10 y+25=0$.
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 $x^2+y^2-2 a x+2 a y+a^2=0$.
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is $x^2+y^2-10 x+10 y+25=0$.
Hence, the required equation of the circle is $x^2+y^2 \pm 10 x \pm 10 y+25=0$.

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