An isosceles right angled triange is in scribed in the circle `x ^(2) +y ^(2) = r ^(2).` If the coordinates of an end of the hypotenuse are (a,b) the coordinates of the vertex are

Options

A. `(-a,-b)`
B. `(b,-a)`
C. `(b,a)`
D. `(-b,-a)`

Answer

To find the coordinates of the vertex of an isosceles right-angled triangle inscribed in a circle given the coordinates of an end of the hypotenuse, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Circle and Triangle Setup**:
The equation of the circle is given as \( x^2 + y^2 = r^2 \). The center of this circle is at the origin (0, 0). The triangle is inscribed in this circle, and we know that one end of the hypotenuse is at the point \( (a, b) \).

2. **Identify the Properties of the Triangle**:
Since the triangle is isosceles and right-angled, the hypotenuse will be the diameter of the circle. The right angle will be at the vertex opposite the hypotenuse.

3. **Find the Other End of the Hypotenuse**:
Since the hypotenuse \( BC \) is the diameter, the midpoint of \( BC \) is the center of the circle. If one end of the hypotenuse is \( (a, b) \), the other end \( (x_2, y_2) \) can be found using the midpoint formula:
\[ \left( \frac{a + x_2}{2}, \frac{b + y_2}{2} \right) = (0, 0) \]
This gives us two equations:
\[ \frac{a + x_2}{2} = 0 \quad \Rightarrow \quad x_2 = -a \]
\[ \frac{b + y_2}{2} = 0 \quad \Rightarrow \quad y_2 = -b \]
Therefore, the coordinates of the other end of the hypotenuse \( C \) are \( (-a, -b) \).

4. **Determine the Coordinates of the Vertex**:
Let the vertex opposite the hypotenuse be \( (h, k) \). Since the triangle is isosceles, the distances from the vertex to each end of the hypotenuse must be equal:
\[ AB = AC \]
Using the distance formula:
\[ AB = \sqrt{(h - a)^2 + (k - b)^2} \]
\[ AC = \sqrt{(h + a)^2 + (k + b)^2} \]
Setting these equal gives:
\[ (h - a)^2 + (k - b)^2 = (h + a)^2 + (k + b)^2 \]

5. **Expand and Simplify**:
Expanding both sides:
\[ (h^2 - 2ah + a^2 + k^2 - 2bk + b^2) = (h^2 + 2ah + a^2 + k^2 + 2bk + b^2) \]
Canceling \( h^2, a^2, k^2, b^2 \) from both sides leads to:
\[ -2ah - 2bk = 2ah + 2bk \]
Rearranging gives:
\[ 4ah + 4bk = 0 \quad \Rightarrow \quad ah + kb = 0 \]

6. **Find the Relationship Between h and k**:
From the equation \( ah + kb = 0 \), we can express \( h \) in terms of \( k \):
\[ h = -\frac{kb}{a} \]

7. **Use the Right Angle Condition**:
The slopes of \( AB \) and \( AC \) must multiply to -1 since they are perpendicular:
\[ \frac{k - b}{h - a} \cdot \frac{k + b}{h + a} = -1 \]
This leads to a relationship that can be solved to find specific values for \( h \) and \( k \).

8. **Final Coordinates**:
After solving the above equations, we find that the coordinates of the vertex \( (h, k) \) can be expressed as:
\[ (b, -a) \quad \text{or} \quad (-b, a) \]
Thus, the coordinates of the vertex are \( (b, -a) \) or \( (-b, a) \).