O is a fixed point and P any point on a given circle , OP is joined and on OP if a point Q is taken so that the ratio of OP and OQ = `k`, a constant quantity, prove that the locus of Q is a circle.

Answer

To prove that the locus of point Q is a circle given the conditions in the problem, we can follow these steps:

### Step 1: Define the Points
Let:
- O be the fixed point with coordinates \( O(x_1, y_1) \).
- P be any point on the circle with coordinates \( P(x_2, y_2) \).
- The circle is defined by the equation \( x^2 + y^2 = r^2 \) (where \( r \) is the radius of the circle).

### Step 2: Establish the Ratio
We know that the ratio of the lengths \( OP \) and \( OQ \) is constant and equal to \( k \). This can be expressed mathematically as:
\[ \frac{OP}{OQ} = k \]
From this, we can express \( OQ \) in terms of \( OP \):
\[ OQ = \frac{OP}{k} \]

### Step 3: Use Section Formula
Let Q divide OP in the ratio \( k:1 \). According to the section formula, the coordinates of point Q can be expressed as:
\[ Q\left(\frac{kx_2 + x_1}{k + 1}, \frac{ky_2 + y_1}{k + 1}\right) \]
where \( (x_2, y_2) \) are the coordinates of point P.

### Step 4: Substitute P's Coordinates
Since point P lies on the circle, we can express its coordinates in terms of a parameter \( \theta \):
\[ x_2 = r \cos \theta, \quad y_2 = r \sin \theta \]
Substituting these into the coordinates of Q gives:
\[ Q\left(\frac{k(r \cos \theta) + x_1}{k + 1}, \frac{k(r \sin \theta) + y_1}{k + 1}\right) \]

### Step 5: Simplify the Coordinates of Q
Let:
\[ h = \frac{k(r \cos \theta) + x_1}{k + 1}, \quad k = \frac{k(r \sin \theta) + y_1}{k + 1} \]
Now we can express \( h \) and \( k \) in terms of \( \theta \).

### Step 6: Find the Locus of Q
To find the locus of Q, we need to eliminate \( \theta \). We can express \( r \cos \theta \) and \( r \sin \theta \) in terms of \( h \) and \( k \):
1. Rearranging the expressions for \( h \) and \( k \):
\[ (k + 1)h - x_1 = kr \cos \theta \quad \text{and} \quad (k + 1)k - y_1 = kr \sin \theta \]
2. Squaring both equations and adding them:
\[ ((k + 1)h - x_1)^2 + ((k + 1)k - y_1)^2 = (kr)^2 \]

### Step 7: Rearranging to Circle Equation
This will yield an equation of the form:
\[ A(h^2 + k^2) + B(h) + C(k) + D = 0 \]
This is the general form of a circle equation, confirming that the locus of Q is indeed a circle.

### Conclusion
Thus, we have shown that the locus of point Q, given the conditions of the problem, is a circle.

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