1. Find the equation of the circle which touches the positive Y -axis at a distanc 4 units from the origin and cuts of an intercept 6 from the X -axis. Let $$ \mathrm{C}(\mathrm{h}, \mathrm{k}) $$ be the centre of the circle touching Y -axis at N and $$ \mathrm{ON}=4 $$. From C , draw perpendiculars CM and CN on X -axis and Y -axis respective) Then, $$ \mathrm{ON}=\mathrm{CM}=4=\mathrm{k} $$

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To find the equation of the circle that touches the positive Y-axis at a distance of 4 units from the origin and cuts off an intercept of 6 from the X-axis, we can follow these steps:

1. **Identify the center of the circle**: 
   Since the circle touches the positive Y-axis at a distance of 4 units from the origin, the center of the circle $$ C(h, k) $$ must be at $$ (h, k) $$ where $$ h = 4 $$ (the distance from the Y-axis) and $$ k $$ is the y-coordinate of the center.

2. **Determine the y-coordinate**:
   We know that the circle touches the Y-axis, so the radius $$ r $$ of the circle is equal to the x-coordinate of the center, which is $$ h = 4 $$. Therefore, the radius $$ r = 4 $$.

3. **Intercept on the X-axis**:
   The circle cuts off an intercept of 6 from the X-axis. The intercept on the X-axis can be found using the equation of the circle. The general equation of a circle with center $$ (h, k) $$ and radius $$ r $$ is given by:
   $$
   (x - h)^2 + (y - k)^2 = r^2
   $$
   Substituting $$ h = 4 $$ and $$ r = 4 $$:
   $$
   (x - 4)^2 + (y - k)^2 = 16
   $$

To find the intercept on the X-axis, we set $$ y = 0 $$:
   $$
   (x - 4)^2 + (0 - k)^2 = 16
   $$
   This simplifies to:
   $$
   (x - 4)^2 + k^2 = 16
   $$

The X-intercepts occur when $$ (x - 4)^2 = 16 - k^2 $$. The intercepts on the X-axis are given by:
   $$
   x - 4 = \pm \sqrt{16 - k^2}
   $$
   Thus, the X-intercepts are:
   $$
   x = 4 + \sqrt{16 - k^2} \quad \text{and} \quad x = 4 - \sqrt{16 - k^2}
   $$

The distance between these two intercepts is:
   $$
   (4 + \sqrt{16 - k^2}) - (4 - \sqrt{16 - k^2}) = 2\sqrt{16 - k^2}
   $$
   We know this distance is equal to 6:
   $$
   2\sqrt{16 - k^2} = 6
   $$
   Dividing both sides by 2:
   $$
   \sqrt{16 - k^2} = 3
   $$
   Squaring both sides:
   $$
   16 - k^2 = 9
   $$
   Rearranging gives:
   $$
   k^2 = 7 \quad \Rightarrow \quad k = \sqrt{7} \quad \text{(since the center is above the X-axis)}
   $$

4. **Final center and equation**:
   The center of the circle is $$ (4, \sqrt{7}) $$ and the radius is $$ 4 $$. Therefore, the equation of the circle is:
   $$
   (x - 4)^2 + (y - \sqrt{7})^2 = 16
   $$

Thus, the equation of the circle is:
$$
(x - 4)^2 + (y - \sqrt{7})^2 = 16
$$
The equation of the circle is $$(x - 4)^2 + (y - \sqrt{7})^2 = 16$$.

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