Problem 104 - A Complete Resource Book in Mathematics for JEE Main

Question

Assertion: The tangent to the circle \(x^{2}+y^{2}=5\) at the point \((1,-2)\) also touches the circle \(x^{2}+y^{2}-8 x+6 y+\) \(20=0\). Then its point of contact is \((3,-1)\). Reason: The equation of tangent to the circle \(x^{2}+y^{2}+\) \(2 g x+2 f y+c=0\) at the point \(\left(x_{1}, y_{1}\right)\) is \(x x_{1}+y y_{1}+g(x\) \(\left.+x_{1}\right)+f\left(y+y_{1}\right)+c=0 .\)

Step-by-Step Solution

Answer

The assertion is correct. The tangent also touches the second circle at the point \((3, -1)\).

Step 1: Find the tangent to the first circle

The given circle equation is \(x^2 + y^2 = 5\). The point of tangency is \((1, -2)\). The tangent at a point \((x_1, y_1)\) to a circle \(x^2 + y^2 = r^2\) is \(xx_1 + yy_1 = r^2\). Substituting, we get: \(x \cdot 1 + y \cdot (-2) = 5\), which simplifies to \(x - 2y = 5\).

Step 2: Verify the tangent with the second circle and find the point of contact

The second circle equation is \(x^2 + y^2 - 8x + 6y + 20 = 0\). We substitute the tangent \(x - 2y = 5\) into this circle equation.1. Solve for \(y\) in terms of \(x\) using the tangent: \(y = \frac{x - 5}{2}\).2. Substitute \(y = \frac{x - 5}{2}\) into the circle equation.3. Solve the resulting quadratic equation to find \(x\). After simplification and solving, we find \(x = 3\), and consequently \(y = -1\) by substituting back into the tangent equation.4. Thus, the point of contact is \((3, -1)\).

Step 3: Confirm assertion and reason

The assertion that the point \((3, -1)\) is the point of contact provided the tangent meets the second circle at exactly one point. This is true as shown in Step 2. The reason provided describes how to formulate a tangent to a circle with a given radius using general point \((x_1, y_1)\), which corresponds to our derivation in Step 1.

Key Concepts

Tangent Line Equation

The tangent line equation for a circle provides a way to determine a straight line that touches the circle at exactly one point, called the point of tangency. For a given circle centered at the origin with the equation \(x^2 + y^2 = r^2\), the tangent at a specific point \((x_1, y_1)\) on the circle can be expressed as follows:

• The equation is \(xx_1 + yy_1 = r^2\).
• This formula results from the geometric property that the radius of a circle and a tangent at a point are perpendicular.
• Simply substitute \((x_1, y_1)\) and \(r\), the radius, to find the tangent line's equation.

This setup is straightforward but crucial for determining the relationship between a circle and a line. It ensures that the tangent touches the circle at one specific point and nowhere else, maintaining a perpendicular relationship with the radius at that point.

Point of Tangency

The point of tangency is where the tangent line and the circle meet. In the context of our exercise, the circle is described by \(x^2 + y^2 = 5\), and the point of tangency given is \((1, -2)\). Let's delve into the significance of this point:

• A tangent's contact point is singular, meaning the tangent does not intersect the circle elsewhere.
• At this point, the tangent line does not cut through the circle but merely touches it, hence the name 'tangent.'
• Knowing this point allows us to precisely formulate the tangent line's equation based on the circle's equation.
• In our exercise, substituting \((1, -2)\) into the tangent equation simplifies to \(x - 2y = 5\), verifying the geometric relationship between the point, the line, and the circle.

The ability to determine this point is crucial in problems involving tangents and ensures a firm understanding of how circles and lines interact geometrically.

Intersection with Second Circle

To determine where a tangent line intersects another circle, one must substitute the tangent's equation into the second circle's equation. This shows if and where the line touches the second circle. Our exercise considers the tangent \(x - 2y = 5\) and the circle \(x^2 + y^2 - 8x + 6y + 20 = 0\):

• First, solve the tangent equation for one variable, such as \(y\), resulting in \(y = \frac{x - 5}{2}\).
• Substitute \(y\) into the second circle's equation, transforming it into a quadratic equation in terms of \(x\).
• Solving this quadratic equation provides the \(x\)-coordinate of the intersection point.
• Following substitution back into \(y = \frac{x - 5}{2}\), we find \(y\) and conclude with the contact point \((3, -1)\).

This process confirms the assertion in the problem that the tangent intersects the second circle at a distinct point, highlighting the importance of carefully solving algebraic systems to determine points of intersection in tangency problems.