the locus of centres of all circles which touch the line x=2a and cut the circle x^2+y^2=a^2 orthogonally is

To find the locus of centers of all circles that touch the line \( x = 2a \) and cut the circle defined by \( x^2 + y^2 = a^2 \) orthogonally, we need to delve into some geometry and algebra. Let’s break this down step by step.

Understanding the Problem

We are dealing with two geometric entities: a line and a circle. The line \( x = 2a \) is a vertical line located at \( x = 2a \) on the Cartesian plane, while the circle \( x^2 + y^2 = a^2 \) is centered at the origin \( (0, 0) \) with a radius of \( a \).

Circle Touching the Line

A circle that touches the line \( x = 2a \) must have its center at a distance equal to its radius from the line. If we denote the center of the circle as \( (h, k) \) and its radius as \( r \), the condition for the circle to touch the line can be expressed as:

• |h - 2a| = r

Circle Cutting the Given Circle Orthogonally

When two circles intersect orthogonally, the relationship between their centers and radii can be described using the following equation. If we have two circles with centers \( (h_1, k_1) \) and \( (h_2, k_2) \) and radii \( r_1 \) and \( r_2 \), they cut orthogonally if:

• \( (h_1 - h_2)^2 + (k_1 - k_2)^2 = r_1^2 + r_2^2 \)

In our case, we have the circle \( x^2 + y^2 = a^2 \) (with center \( (0, 0) \) and radius \( a \)) and our unknown circle with center \( (h, k) \) and radius \( r \). The orthogonality condition becomes:

• \( h^2 + k^2 = r^2 + a^2 \)

Combining Conditions

Now, we have two conditions to work with:

• |h - 2a| = r
• h^2 + k^2 = r^2 + a^2

Substituting for r

From the first equation, we can express \( r \) as:

• r = |h - 2a|

Now, we need to substitute \( r \) back into the second equation. We have:

• h^2 + k^2 = (|h - 2a|)^2 + a^2

Expanding and Simplifying

Let’s square the absolute value:

• h^2 + k^2 = (h - 2a)^2 + a^2
• h^2 + k^2 = h^2 - 4ah + 4a^2 + a^2
• h^2 + k^2 = h^2 - 4ah + 5a^2

Now, we can eliminate \( h^2 \) from both sides:

• k^2 = -4ah + 5a^2

Finding the Locus

The equation we derived, \( k^2 = -4ah + 5a^2 \), represents the locus of centers of all circles that meet the specified conditions. To express it in a more recognizable format, we can rearrange it:

• k^2 + 4ah - 5a^2 = 0

Identifying the Nature of the Locus

This equation is quadratic in terms of \( k \), indicating that the locus is a parabola that opens vertically. The vertex of this parabola can be found by completing the square or using the vertex formula for parabolas. The shape and orientation are determined by the coefficients present in the equation.

Final Thoughts

In summary, the locus of the centers of all circles that touch the line \( x = 2a \) and cut the circle \( x^2 + y^2 = a^2 \) orthogonally is given by the equation \( k^2 + 4ah - 5a^2 = 0 \). This represents a parabola in the Cartesian plane. Understanding these relationships between geometric shapes is fundamental in coordinate geometry and helps reinforce the connection between algebra and geometry.