The straight line px+qy=1 makes the `ax^2+2hxy+by^2=c`, a chord which subtends a right angle at the origin . Show that `c(p^2+q^2)`=a+b . Also show that px+qy=1 touches a circle whose radius is `sqrt(c/(a+b))`

Answer

To solve the problem, we need to show two things:

1. That \( c(p^2 + q^2) = a + b \).
2. That the line \( px + qy = 1 \) touches a circle whose radius is \( \sqrt{\frac{c}{a + b}} \).

Let's go through the solution step by step.

### Step 1: Find the points of intersection of the line with the axes

The line given is \( px + qy = 1 \).

- **Intersection with the x-axis**: Set \( y = 0 \).
\[ px = 1 \implies x = \frac{1}{p} \implies \text{Point A} = \left(\frac{1}{p}, 0\right) \]

- **Intersection with the y-axis**: Set \( x = 0 \).
\[ qy = 1 \implies y = \frac{1}{q} \implies \text{Point B} = \left(0, \frac{1}{q}\right) \]

### Step 2: Substitute points into the curve equation

The curve is given by:
\[ ax^2 + 2hxy + by^2 = c \]

We need to check that both points A and B lie on the curve.

- For Point A \( \left(\frac{1}{p}, 0\right) \):
\[ a\left(\frac{1}{p}\right)^2 + 2h\left(\frac{1}{p}\right)(0) + b(0)^2 = c \]
This simplifies to:
\[ \frac{a}{p^2} = c \implies a = cp^2 \quad (1) \]

- For Point B \( \left(0, \frac{1}{q}\right) \):
\[ a(0)^2 + 2h(0)\left(\frac{1}{q}\right) + b\left(\frac{1}{q}\right)^2 = c \]
This simplifies to:
\[ \frac{b}{q^2} = c \implies b = cq^2 \quad (2) \]

### Step 3: Add equations (1) and (2)

Now we can add the two equations:
\[ a + b = cp^2 + cq^2 = c(p^2 + q^2) \]
Thus, we have shown that:
\[ c(p^2 + q^2) = a + b \]

### Step 4: Show that the line touches a circle

To show that the line \( px + qy = 1 \) touches a circle, we need to find the radius of the circle and the distance from the center of the circle (which is at the origin) to the line.

The formula for the distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line \( px + qy - 1 = 0 \), we have:
- \( A = p \)
- \( B = q \)
- \( C = -1 \)

Substituting \( (0, 0) \) for \( (x_0, y_0) \):
\[ d = \frac{|p(0) + q(0) - 1|}{\sqrt{p^2 + q^2}} = \frac{1}{\sqrt{p^2 + q^2}} \]

### Step 5: Relate the distance to the radius

The radius \( r \) of the circle is given by:
\[ r = \sqrt{\frac{c}{a + b}} \]

From Step 3, we know that \( a + b = c(p^2 + q^2) \). Therefore:
\[ r = \sqrt{\frac{c}{c(p^2 + q^2)}} = \sqrt{\frac{1}{p^2 + q^2}} \]

### Conclusion

Since the distance from the origin to the line is \( \frac{1}{\sqrt{p^2 + q^2}} \) and the radius of the circle is also \( \sqrt{\frac{1}{p^2 + q^2}} \), we conclude that the line touches the circle.