The equation of circle which touches x y axis and whose perpendicular distance of centre of circle from 3 x+4 y+11=0 is 5 is (Given that circle lies in lst quadrant)

Question

The equation of circle which touches x y axis and whose perpendicular distance of centre of circle from 3 x+4 y+11=0 is 5 is (Given that circle lies in lst quadrant) (A) x2+y2+4 x+4 y+4=0 (B) x2+y2-4 x-4 y+4=0 (C) x2+y2-4 x-4 y+8=0 (D) x2+y2-4 x-4 y-4=0

Tardigrade Admin · Accepted Answer

Let centre be (α, α) α>0 ⇒ 5=|(3 α+4 α+11/5)| ⇒ 7 α+11=± 25 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/a687ef99ada416043404b9d38e39120d-.png" /> 7 α=14,7 α=-36 α=2, α=(-36/7) so circle (x-2)2+(y-2)2=4

Q. The equation of circle which touches $x & y$ axis and whose perpendicular distance of centre of circle from $3 x + 4 y + 11 = 0$ is 5 is (Given that circle lies in lst quadrant)

$x^{2} + y^{2} + 4 x + 4 y + 4 = 0$

$x^{2} + y^{2} - 4 x - 4 y + 4 = 0$

$x^{2} + y^{2} - 4 x - 4 y + 8 = 0$

$x^{2} + y^{2} - 4 x - 4 y - 4 = 0$

Let centre be $(α, α) & α > 0$
$\Rightarrow 5 = ∣ ∣ \frac{3 α + 4 α + 11}{5} ∣ ∣$
$\Rightarrow 7 α + 11 = \pm 25$

$7 α = 14, 7 α = - 36$
$α = 2, α = \frac{- 36}{7}$
so circle $(x - 2)^{2} + (y - 2)^{2} = 4$

Q. The equation of circle which touches x&y axis and whose perpendicular distance of centre of circle from 3x+4y+11=0 is 5 is (Given that circle lies in lst quadrant)

x2+y2+4x+4y+4=0

x2+y2−4x−4y+4=0

x2+y2−4x−4y+8=0

x2+y2−4x−4y−4=0