Question

Circle(s) touching x-axis at a distance 3 from the origin and having an intercept of length $2\sqrt{7}$ on y-axis is (are)

• A. $x^2+y^2-6x+8y+9=0$
• B. $x^2+y^2-6x+7y+9=0$
• C. $x^2+y^2-6x-8y+9=0$
• D. $x^2+y^2-6x-7y+9=0$

Answer 1

Answer: (A), (C)

$x^2+y^2-6x+8y+9=0$
$x^2+y^2-6x-8y+9=0$

a trianlge gets formed, with sides x-axis $3$, y-axis $\left(\frac{2\sqrt{7}}{2}\right)$ and $k$ we get,

$k^2=3^2+{\left(\frac{2\sqrt{7}}{2}\right)}^2$

$\therefore k=4$

y-axis distance $k=\pm 4$

x-axis distance $h=3$

now,

circle,

$(x-3{)}^2+(y-4{)}^2=4^2$

$x^2+y^2-6x-8y+9=0$

OR

$(x+3{)}^2+(y-4{)}^2=16$

$x^2+y^2+6x-8y+9=0$

OR

$(x-3{)}^2+(y+4{)}^2=16$