Question

The equation of the circle passing through the point \( (1, 2) \) and touching both axes is:

• A. \( (x - 1)^2 + (y - 2)^2 = 1 \)
• B. \( (x - 2)^2 + (y - 1)^2 = 1 \)
• C. \( (x - 1)^2 + (y - 1)^2 = 1 \)
• D. \( (x - 2)^2 + (y - 2)^2 = 4 \)

Hint:

When a circle touches both coordinate axes, the center lies at \( (r, r) \), and the equation becomes \( (x - r)^2 + (y - r)^2 = r^2 \).

Answer 1

The Correct Option is D

We are given that the circle touches both the x-axis and y-axis.
This implies that the radius \( r \) of the circle is equal to the x and y coordinates of the center (since distance from the center to each axis = radius).
Let the center be at \( (a, a) \) so the equation becomes:
\[ (x - a)^2 + (y - a)^2 = a^2 \] Given that the circle passes through the point \( (1, 2) \), we substitute into the equation:
\[ (1 - a)^2 + (2 - a)^2 = a^2 \] Expanding and simplifying:
\[ (1 - a)^2 = a^2 - 2a + 1\\ (2 - a)^2 = a^2 - 4a + 4\\ \Rightarrow (1 - a)^2 + (2 - a)^2 = a^2\\ \Rightarrow a^2 - 2a + 1 + a^2 - 4a + 4 = a^2\\ \Rightarrow 2a^2 - 6a + 5 = a^2\\ \Rightarrow a^2 - 6a + 5 = 0 \] Solving the quadratic equation:
\[ a^2 - 6a + 5 = 0 \Rightarrow a = 5 \text{ or } a = 1 \] Let's check both values:
1. If \( a = 5 \):
Center = (5, 5), Radius = 5
Equation: \( (x - 5)^2 + (y - 5)^2 = 25 \)
Check if it passes through (1, 2):
\[ (1 - 5)^2 + (2 - 5)^2 = 16 + 9 = 25 \Rightarrow \text{Yes, satisfies} \] 2. If \( a = 1 \):
Center = (1, 1), Radius = 1
Equation: \( (x - 1)^2 + (y - 1)^2 = 1 \)
Check if it passes through (1, 2):
\[ (1 - 1)^2 + (2 - 1)^2 = 0 + 1 = 1 \Rightarrow \text{Yes, satisfies} \] So, both values of \( a = 1 \) and \( a = 5 \) satisfy the condition. Therefore, there are two such circles.