The general equation of a circle is:

$ (x-h)^2 + (y-k)^2 = r^2 $

Here, $(h, k)$ represents the center of the circle, $r$ is its radius, and $(x, y)$ is any point on the circle.

Given that the circle's center lies on the line $y = x + 1$, we have $k = h + 1$.

We can substitute this relationship and use the following two points through which the circle passes: $(-2, 1)$ and $(0, 3)$.

By setting the distances equal, we obtain:

$ (-2-h)^2 + (1-(h+1))^2 = (0-h)^2 + (3-(h+1))^2 $

Simplifying, we have:

$ (2+h)^2 + h^2 = h^2 + (2-h)^2 $

This simplifies to:

$ (2+h)^2 = (2-h)^2 $

Solving for $h$, we find:

$ h = 0, \quad k = 1 $

The circle's center is $(0, 1)$ with radius $2$ (the same as the distance from $(0, 1)$ to $(0, 3)$).

Next, we verify which given points lie on this circle:

Observing the point $(\sqrt{2}+1, 1)$, it results in an irrational distance, hence it can be dismissed. Similarly, $(1, \sqrt{3}-1)$ results in an irrational value, ruling it out.

Checking the remaining options:

$ \begin{align*} (1, \sqrt{2}+1): &\quad 1^2 + (\sqrt{2} + 1 - 1)^2 = 1 + (\sqrt{2})^2 = 1 + 2 = 3 \quad (\neq 4) \\ (1, \sqrt{3}+1): &\quad 1^2 + (\sqrt{3} + 1 - 1)^2 = 1 + (\sqrt{3})^2 = 1 + 3 = 4 \end{align*} $

Thus, the point $(1, \sqrt{3}+1)$ lies on the circle.

Algebra

Trigonometry

Calculus

Coordinate Geometry

Circle · Mathematics · IAT (IISER)

MCQ (Single Correct Answer)