Question:

Circle(s) touching X-axis at a distance 3 from the origin and having an intercept of length $2 \sqrt7$ on Y-axis is/are

Updated On: Jun 14, 2022
  • $x^2 + y2 - 6x + 8y + 9 = 0$
  • $x^2 + y2 - 6x + 7y + 9 = 0$
  • $x^2 + y2 - 6x - 8y + 9 = 0$
  • $x^2 + y2 - 6x - 7y + 9 = 0$
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The Correct Option is C

Solution and Explanation

PLAM
Here, the length of intercept on 7-axis is $\Rightarrow \, \, 2 \sqrt{ f^2} - c$
and if circle touches X-axis
$\Rightarrow \hspace12mm g^2 = c$
for $x^2 + y^2 + 2gx + 2fy + c = 0$
Here, $ \hspace5mm x^2 + y^2 + 2gx + 2fy + c = 0$
passes through (3,0).
$\Rightarrow \hspace12mm 9 + 6 g + c = 0 \hspace8mm ...(i)$
$ \hspace20mm g^2 = c \hspace15mm ...(ii)$
and $ \hspace15mm 2 \sqrt {f^2- c} = 2 \sqrt7 $
$\hspace15mm f^2 - c = 7 \hspace15mm ...(iii)$
From Eqs. (i) and (ii), we get
$\hspace15mm g^2 + 6g + 9 = 0$
$\Rightarrow \hspace15mm (g + 3)^2 = 0$
$\Rightarrow \hspace15mm (g = - 3) and c = 9$
$\therefore \hspace15mm f^2 = 16 \Rightarrow f = \pm 4$
$\therefore \hspace15mm x^2 + y^2 - 6x \pm 8y + 9 = 0 $
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