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 \( \sqrt{\frac{c}{a + b}} \).

### Step 1: Convert the given line and curve into a homogeneous equation

We start with the line given by:

\[ px + qy = 1 \]

We can rewrite this as:

\[ \frac{px + qy}{1} = 1 \]

Now, we substitute this into the curve equation \( ax^2 + 2hxy + by^2 = c \).

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

We substitute \( px + qy = 1 \) into the curve equation:

\[ ax^2 + 2hxy + by^2 - c(px + qy)^2 = 0 \]

Expanding \( (px + qy)^2 \):

\[ (px + qy)^2 = p^2x^2 + 2pqxy + q^2y^2 \]

So, we have:

\[ ax^2 + 2hxy + by^2 - c(p^2x^2 + 2pqxy + q^2y^2) = 0 \]

### Step 3: Rearranging the equation

Rearranging gives us:

\[ (ax^2 - cp^2x^2) + (by^2 - cq^2y^2) + (2hxy - 2cpqxy) = 0 \]

This simplifies to:

\[ (a - cp^2)x^2 + (b - cq^2)y^2 + (2h - 2cpq)xy = 0 \]

### Step 4: Condition for the chord to subtend a right angle at the origin

For the chord to subtend a right angle at the origin, the following condition must hold:

\[ (a - cp^2) + (b - cq^2) = 0 \]

This leads to:

\[ a + b = c(p^2 + q^2) \]

Thus, we have shown that:

\[ c(p^2 + q^2) = a + b \]

### Step 5: Show that the line touches a circle

Next, we need to show that the line \( px + qy = 1 \) touches a circle with radius \( \sqrt{\frac{c}{a + b}} \).

The general form of the equation of a circle is:

\[ x^2 + y^2 = r^2 \]

The radius \( r \) can be expressed in terms of the coefficients of the line and the circle. The distance from the center of the circle (which we can take as the origin) to the line \( px + qy = 1 \) is given by:

\[ \text{Distance} = \frac{|px + qy - 1|}{\sqrt{p^2 + q^2}} \]

### Step 6: Set the distance equal to the radius

For the line to touch the circle, this distance must equal the radius. Setting \( h = 0 \) and \( k = 0 \) (the center of the circle at the origin), we have:

\[ \frac{|1|}{\sqrt{p^2 + q^2}} = \sqrt{\frac{c}{a + b}} \]

Squaring both sides gives:

\[ 1 = \frac{c}{a + b}(p^2 + q^2) \]

Thus, we have:

\[ \sqrt{\frac{c}{a + b}} = \frac{1}{\sqrt{p^2 + q^2}} \]

This confirms that the line \( px + qy = 1 \) indeed touches the circle with radius \( \sqrt{\frac{c}{a + b}} \).

### Final Result

We have shown both parts of the problem:

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