Question

Let A and B are two points on x-axis and y-axis respectively and circle passes through A, B and O (origin). Tangent at origin is at a distance of 3 unit and 5 unit from point A and B respectively, then radius of circle is

Answer 1

4

Answer 2

Step 1. Let the circle have centre $(h,k)$ and radius $r$. Since the circle passes through the origin $O(0,0)$,

$$h^2 + k^2 = r^2.$$

Step 2. Tangent at $O$ is perpendicular to radius $O C$ with direction vector $(h,k)$. Thus the tangent line is

$$hx + ky = 0.$$

Distance from $A(a,0)$ to this line is

$$\frac{|h\,a|}{\sqrt{h^2+k^2}} = \frac{|h\,a|}{r} = 3 \quad\Longrightarrow\quad |h\,a| = 3r.$$

Similarly, distance from $B(0,b)$ is

$$\frac{|k\,b|}{r} = 5 \quad\Longrightarrow\quad |k\,b| = 5r.$$

Step 3. Since $A$ and $B$ lie on the circle:

$$(a-h)^2 + (0-k)^2 = r^2 \quad\Longrightarrow\quad (a-h)^2 + k^2 = r^2 = h^2+k^2 \quad\Longrightarrow\quad a-h = \pm h.$$

Choosing $a-h = h$ (so that $a\neq0$) gives $a = 2h$.
Similarly for $B$:

$$h^2 + (b-k)^2 = r^2 \quad\Longrightarrow\quad b-k = \pm k \quad\Longrightarrow\quad b = 2k.$$

Step 4. Combine with distances:

$$2h = a = \frac{3r}{h} \quad\Longrightarrow\quad 2h^2 = 3r,$$ $$2k = b = \frac{5r}{k} \quad\Longrightarrow\quad 2k^2 = 5r.$$

Hence,

$$h^2 = \frac{3r}{2}, \quad k^2 = \frac{5r}{2} \quad\Longrightarrow\quad h^2 + k^2 = \frac{3r+5r}{2} = 4r.$$

But $h^2+k^2 = r^2$, so $r^2 = 4r$ and $r = 4$ (reject $r=0$).