The equation of a circle passing through (3,-6) and touching both the axes is

Question

The equation of a circle passing through (3,-6) and touching both the axes is (A) x2+y2-6 x+6 y+9=0 (B) x2+y2+6 x-6 y+9=0 (C) x2+y2+30 x-30 y+225=0 (D) x2+y2+30 x+30 y+225=0

Tardigrade Admin · Accepted Answer

Now (r-3)2+(-r+6)2=r2 r2-18 r+45=0 ⇒ r=3,15 Hence circle <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/5845dd11dcdfe4d6d1628529733c4337-.png" /> (x-3)2+(y+3)2=32 x2+y2-6 x+6 y+9=0 (x-15)2+(y+15)2=(15)2 ⇒ x2+y2-30 x+30 y+225=0

Q. The equation of a circle passing through $(3, - 6)$ and touching both the axes is

$x^{2} + y^{2} - 6 x + 6 y + 9 = 0$

$x^{2} + y^{2} + 6 x - 6 y + 9 = 0$

$x^{2} + y^{2} + 30 x - 30 y + 225 = 0$

$x^{2} + y^{2} + 30 x + 30 y + 225 = 0$

Now
$(r - 3)^{2} + (- r + 6)^{2} = r^{2}$
$r^{2} - 18 r + 45 = 0$
$\Rightarrow r = 3, 15$
Hence circle

$(x - 3)^{2} + (y + 3)^{2} = 3^{2}$
$x^{2} + y^{2} - 6 x + 6 y + 9 = 0$
$(x - 15)^{2} + (y + 15)^{2} = (15)^{2}$
$\Rightarrow x^{2} + y^{2} - 30 x + 30 y + 225 = 0$

Q. The equation of a circle passing through (3,−6) and touching both the axes is

x2+y2−6x+6y+9=0

x2+y2+6x−6y+9=0

x2+y2+30x−30y+225=0

x2+y2+30x+30y+225=0