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, \( \mathrm{ON}=\mathrm{CM}=4=\mathrm{k} \)

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: \[ (x - 4)^2 + (y - \sqrt{7})^2 = 16 \]