The equation of the circle passing through the point (2,1) and touching y-axis at the origin is

Question

The equation of the circle passing through the point (2,1) and touching y-axis at the origin is (A) x2+y2-5 x=0 (B) 2 x2+2 y2-5 x=0 (C) x2+y2+5 x=0 (D) x2-y2-5 x=0

Tardigrade Admin · Accepted Answer

We have the equation of circle x2+y2+2 g x+2 f y+c=0 But it passes through (0,0) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/d8a16a013d2332abcf2f583a9ad44663-.png" /> and (2,1), then c=0 .....(i) 5+4 g+2 f=0.....(ii) As it touches y-axis ∴ 2 √ f 2- c =0 [∵ c =0] or f=0, from (ii),g=-(5/4)

Q. The equation of the circle passing through the point $(2, 1)$ and touching $y$ -axis at the origin is

$x^{2} + y^{2} - 5 x = 0$

$2 x^{2} + 2 y^{2} - 5 x = 0$

$x^{2} + y^{2} + 5 x = 0$

$x^{2} - y^{2} - 5 x = 0$

We have the equation of circle
$x^{2} + y^{2} + 2 gx + 2 f y + c = 0$
But it passes through $(0, 0)$

and $(2, 1)$ , then
$c = 0.....$ (i)
$5 + 4 g + 2 f = 0.....$ (ii)
As it touches $y$ -axis
$∴ 2 f^{2} - c = 0 [∵ c = 0]$
or $f = 0$ , from (ii), $g = - \frac{5}{4}$

Q. The equation of the circle passing through the point (2,1) and touching y-axis at the origin is

x2+y2−5x=0

2x2+2y2−5x=0

x2+y2+5x=0

x2−y2−5x=0