The locus of the centre of a circle which touches externally the circle x^2+y^2 -6x-6y+14=0 and also touches the y axis is given by the equation is

To find the locus of the center of a circle that touches both the given circle and the y-axis, we need to break down the problem step by step. Let's start by analyzing the given circle's equation and then derive the required locus.

Understanding the Given Circle

The equation of the circle is:

x² + y² - 6x - 6y + 14 = 0

We can rewrite this in standard form by completing the square.

Completing the Square

• For x: x² - 6x can be rewritten as (x - 3)² - 9.
• For y: y² - 6y can be rewritten as (y - 3)² - 9.

Substituting these back into the equation gives:

(x - 3)² - 9 + (y - 3)² - 9 + 14 = 0

Which simplifies to:

(x - 3)² + (y - 3)² = 4

Circle Touching the Y-Axis

Next, we need to consider the condition that our new circle touches the y-axis. If a circle touches the y-axis, the distance from its center to the y-axis must equal its radius. Let’s denote the center of our new circle as (h, k) and its radius as r.

Distance to the Y-Axis

The distance from the center (h, k) to the y-axis is simply |h|. Therefore, for the circle to touch the y-axis, we have:

|h| = r

Touching the Given Circle Externally

Now, for our new circle to touch the given circle externally, the distance between their centers must equal the sum of their radii. The center of the given circle is (3, 3) with radius 2, and the center of our new circle is (h, k) with radius r.

Distance Between Centers

The distance between the centers (h, k) and (3, 3) is given by:

√[(h - 3)² + (k - 3)²]

For external tangency, we set this equal to the sum of the radii:

√[(h - 3)² + (k - 3)²] = r + 2

Combining the Conditions

Now we have two equations:

• |h| = r
• √[(h - 3)² + (k - 3)²] = r + 2

Substituting the first equation into the second gives us:

√[(h - 3)² + (k - 3)²] = |h| + 2

Squaring Both Sides

To eliminate the square root, we square both sides:

(h - 3)² + (k - 3)² = (|h| + 2)²

Expanding both sides leads to:

(h - 3)² + (k - 3)² = h² + 4|h| + 4

Final Steps to Derive the Locus

Now, we can simplify and rearrange this equation to find the locus of (h, k). This will involve some algebraic manipulation, but ultimately, we will arrive at a relationship between h and k that describes the locus.

Resulting Equation

After simplifying, we find that the locus of the center of the circle that touches the given circle externally and the y-axis is:

k² - 6k + h² - 6h + 9 = 0

or, more simply, it can be expressed as:

k = 3 ± √(h² + 4|h| - 4)

This equation describes a parabola, which is the locus of the centers of all such circles. Thus, the final answer is the equation representing this locus.