Question

The equation of the circle which touches the axis of y at a distance $+4$ from the origin and cuts off an intercept $6$ from the $+ve$ direction of x-axis is

• A. $x^2+y^2-10x\pm 8y-16=0$
• B. $x^2+y^2+10x\pm 8y+16=0$
• C. $x^2+y^2-10x\pm 8y+16=0$
• D. $x^2+y^2-8x\pm 10y-16=0$

Answer 1

The correct option is C $x^2+y^2-10x\pm 8y+16=0$

For a circle with center (a,b) and radius r,

Equation is $(x-a{)}^2+(y-b{)}^2=r^2$

There can be two scenarios for the given conditions as depicted in above figures.

For (A),

In $△OMN$, using pythagoras theorem,

$ON=\sqrt{O{M}^2+M{N}^2}=\sqrt{4^2+3^2}=5$

So, radius of circle = 5.

And, center = (5,4)

Thus, Equation of cricle

$\to$ $(x-5{)}^2+(y-4{)}^2=5^2$

$\to$ $x^2+y^2-10x-8y+16=0$

Similarly, For (B).

Equation of circle,

$\to$ $(x-5{)}^2+(y+4{)}^2=5^2$

$\to$ $x^2+y^2-10x+8y+16=0$