Question

The circle passing through the point \((-1,0)\) and touching the \(y\) -axis at \((0,2)\) also passes through the point (A) \(\left(-\frac{3}{2}, 0\right)\) (B) \(\left(-\frac{5}{2}, 2\right)\) (C) \(\left(-\frac{3}{2}, \frac{5}{2}\right)\) (D) \((-4,0)\)

Short answer

Point (C) \( \left(-\frac{3}{2}, \frac{5}{2}\right) \) is on the circle.

Step-by-step solution

Determine the radius of the circle

Since the circle touches the y-axis at (0,2), the radius of the circle is equal to the distance from the center to this point of tangency. The center must be directly left of (0,2) on the line x = -1 because the circle also passes through (-1,0). The distance from (-1,0) to (0,2), which is the length of the radius, can be calculated using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). So, \( r = \sqrt{(0 - (-1))^2 + (2 - 0)^2} = \sqrt{1 + 4} = \sqrt{5} \).

Find the coordinates of the center

The center of the circle has the same y-coordinate as the point of tangency (0,2) but is \( \sqrt{5} \) units to the left since it must also pass through (-1,0). So, the center is \( (-1 - \sqrt{5}, 2) \) since it is \( \sqrt{5} \) units away from (-1,0) to the left.

Use the circle equation to check which point lies on the circle

The general equation of a circle with center \( (h,k) \) and radius \( r \) is \( (x-h)^2 + (y-k)^2 = r^2 \). For this circle equation with center \( (-1 - \sqrt{5},2) \) and radius \( \sqrt{5} \), we have \( (x + 1 + \sqrt{5})^2 + (y - 2)^2 = 5 \). Substitute each point given in options (A) to (D) to see which satisfies the equation.

Verify Option (A)

Substitute \( (-\frac{3}{2}, 0) \) into the circle equation: \( (x + 1 + \sqrt{5})^2 + (y - 2)^2 \). We get \( (-\frac{3}{2} + 1 + \sqrt{5})^2 + (0 - 2)^2 = (-\frac{1}{2} + \sqrt{5})^2 + 4 \). This does not simplify to 5, hence point (A) is not on the circle.

Verify Option (B)

Substitute \( (-\frac{5}{2}, 2) \) into the circle equation: \( (-\frac{5}{2} + 1 + \sqrt{5})^2 + (2 - 2)^2 \). Simplifying \( (-\frac{3}{2} + \sqrt{5})^2 + 0 \), which does not equal 5, so point (B) is also not on the circle.

Verify Option (C)

Substitute \( (-\frac{3}{2}, \frac{5}{2}) \) into the circle equation: \( (-\frac{3}{2} + 1 + \sqrt{5})^2 + (\frac{5}{2} - 2)^2 \) which simplifies to \( (-\frac{1}{2} + \sqrt{5})^2 + (\frac{1}{2})^2 \) and this equals 5. Hence point (C) is on the circle.

Verify Option (D)

Substitute \( (-4, 0) \) into the circle equation: \( (-4 + 1 + \sqrt{5})^2 + (0 - 2)^2 \) which simplifies to \( (-3 + \sqrt{5})^2 + 4 \) which does not equal 5, so point (D) is not on the circle.

Key concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a branch of mathematics that involves the study of geometric shapes and figures using a coordinate system. This approach allows for the examination of the properties of these shapes through algebraic equations. In the context of the JEE Advanced mathematics problem involving a circle, we utilize coordinate geometry to define the position of the circle, its radius, and also to determine the position of points and tangents relative to the circle.

When approaching circle equations in coordinate geometry, two crucial pieces of information are typically needed: the coordinates of the center of the circle and the length of its radius. Once these are known, the entire circle's position can be clearly defined in the coordinate plane. This is particularly powerful when solving problems where circles interact with other geometric entities, such as lines or other circles.

Equation of a Circle

The equation of a circle is a fundamental concept in geometry, which serves as a bridge connecting algebraic expressions with geometric figures. In its standard form, the equation is given as \( (x-h)^2 + (y-k)^2 = r^2 \), where \( (h,k) \) denote the coordinates of the center of the circle, and \( r \) is its radius. This algebraic equation beautifully encapsulates the definition of a circle: the set of all points in a plane that are a fixed distance \( (r) \) from a given point \( (the center) \).

Solving circle problems often involves plugging coordinates of given points into the circle equation to test whether those points lie on the circle—that is, whether they satisfy the equation. This step was demonstrated in the solution steps verifying options \( (A), (B), (C), \) and \( (D) \) to identify which point, if any, also lies on the circle in question.

Tangent to a Circle

A tangent to a circle is a straight line that touches the circle at exactly one point. This point of contact is known as the point of tangency. A key property of a tangent to a circle is that it is perpendicular to the radius of the circle at the point of tangency. In the context of JEE Advanced problems, understanding the tangent's behavior is essential in determining the circle's position and solving the problems that involve tangency conditions.

For instance, if a circle is tangent to an axis of a coordinate system at a known point, this tangency condition can help calculate the circle's radius and, subsequently, its center. This method was used in the provided steps to solve the circle problem, where the circle was known to touch the y-axis at \( (0,2) \) and also to pass through another point, which together allowed for the determination of both the center and the radius of the circle.

JEE Advanced Mathematics

JEE Advanced Mathematics covers a wide range of topics, including algebra, calculus, trigonometry, and coordinate geometry. These subjects are crucial for the JEE Advanced aspirants as they form the basis of the questions posed in the examination. The intricate problems require a deep understanding of core concepts and the ability to apply them to solve complex problems efficiently.

The exercise presented here exemplifies the kind of geometric application of algebraic principles that students can expect in the JEE Advanced examination. By converting abstract geometric conditions into algebraic equations, students are tested on their analytic skills and their conceptual clarity. Regular practice with these types of questions, including understanding tangents, circle equations, and coordinate systems, is essential for success in JEE Advanced mathematics.