Find the equation of circle which touches $x$ & $y$ axis & perpendicular distance of center of circle from $3 x + 4 y + 11 = 0$ is $5$ . Given circle lies $1^{s t}$ quadrant

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Solution

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Solution -
Radius = h.
then $⊥$ distance on $3 x + 4 y + 11 = 0$
from centre is 5.
$∣ ∣ ∣ \frac{3 h + 4 h + 11}{5} ∣ ∣ ∣ = 5$
$7 h + 11 = 25$
$h = 2$
circle - $(x - 2)^{2} + (y - 2)^{2} = 2^{2}$
$(x - 2)^{2} + (y - 2)^{2} = 4$

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