Find the locus of the feet of the perpendiculars drawn from the point (b, 0) on tangents to the circle `x^(2) + y^(2) =a^(2)`

Answer

To find the locus of the feet of the perpendiculars drawn from the point (b, 0) on tangents to the circle defined by the equation \(x^2 + y^2 = a^2\), we can follow these steps:

### Step 1: Understand the Circle and Tangent Equation
The given circle is centered at the origin (0, 0) with a radius of \(a\). The general equation of the tangent to the circle at any point can be expressed as:
\[ y = mx \pm a\sqrt{1 + m^2} \]
where \(m\) is the slope of the tangent.

### Step 2: Define the Point and the Feet of the Perpendicular
Let \(P(b, 0)\) be the point from which the perpendicular is drawn to the tangent. Let \(H(h, k)\) be the foot of the perpendicular from point \(P\) to the tangent.

### Step 3: Find the Slope of the Perpendicular
The slope of the line \(PH\) (from \(P\) to \(H\)) can be calculated as:
\[ \text{slope of } PH = \frac{k - 0}{h - b} = \frac{k}{h - b} \]
Since \(PH\) is perpendicular to the tangent, the slope of the tangent \(m\) is related to the slope of \(PH\) as follows:
\[ \frac{k}{h - b} = -\frac{1}{m} \quad \Rightarrow \quad k = -\frac{h - b}{m} \]

### Step 4: Substitute into the Tangent Equation
Substituting \(k\) into the tangent equation, we have:
\[ -\frac{h - b}{m} = mh \pm a\sqrt{1 + m^2} \]

### Step 5: Rearranging the Equation
Rearranging gives us:
\[ h - b = -m^2h \mp ma\sqrt{1 + m^2} \]
This can be simplified to express \(h\) in terms of \(b\) and \(m\).

### Step 6: Eliminate \(m\)
To find the locus, we need to eliminate \(m\). We can do this by substituting the expression for \(m\) back into the equation and simplifying.

### Step 7: Replace Variables for Locus
Replace \(h\) with \(x\) and \(k\) with \(y\) in the final equation to express the relationship purely in terms of \(x\) and \(y\).

### Final Equation
After simplification, we arrive at the locus equation:
\[ y^2 - b y + x^2 = a^2 \]