Problem 104 - A Complete Resource Book in Mathematics for JEE Main 2017

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 point of contact is (3, -1).

Step 1: Determine the Tangent Equation to First Circle

The given circle's equation is \(x^2 + y^2 = 5\). We need to find the equation of the tangent at the point \((1, -2)\). The equation of the tangent to a circle \(x^2 + y^2 = r^2\) at \((x_1, y_1)\) is \(xx_1 + yy_1 = r^2\). Substituting the given point and radius squared (5), the equation becomes \(x\cdot1 + y\cdot(-2) = 5\), i.e., \(x - 2y = 5\).

Step 2: Verify if Tangent touches Second Circle

The tangent \(x - 2y = 5\) must also touch the second circle, given by its equation \(x^2 + y^2 - 8x + 6y + 20 = 0\). To check this, rewrite the circle's equation in general form and calculate the perpendicular distance from its center \((4, -3)\) to the tangent line:The line equation \(x - 2y = 5\) can be rearranged to \(x - 2y - 5 = 0\). The circle center \((4, -3)\) is obtained from completing the square. The perpendicular distance is:\[\frac{|4 - 2(-3) - 5|}{\sqrt{1 + (-2)^2}} = \frac{7}{\sqrt{5}}\]Check the radius of the circle by computing the distance \(\sqrt{4^2 + (-3)^2 - 20} = \sqrt{5}\).Since the distance equals the radius, the tangent touches the circle.

Step 3: Find Point of Contact

To find the point of contact, solve the system of equations formed by the circle's equation and the tangent's equation:1. \(x - 2y = 5\)2. \(x^2 + y^2 - 8x + 6y + 20 = 0\)Substitute \(x = 2y + 5\) from the tangent equation into the circle's equation:\((2y + 5)^2 + y^2 - 8(2y + 5) + 6y + 20 = 0\)Simplify and solve for \(y\):\[4y^2 + 20y + 25 + y^2 - 16y - 40 + 6y + 20 = 0\]Combine terms:\[5y^2 + 10y + 5 = 0\]Solve this quadratic for \(y\), yielding \(y = -1\), assuming some simplification steps or recognizing this is apparent. Substitute back to get \(x = 3\):Thus, the point of contact is \((3, -1)\).

Key Concepts

Circle

A circle is one of the most fundamental shapes in geometry. Its definition is simple yet powerful: a set of all points in a plane that are equidistant from a given point, the center. This uniform distance from the center is called the radius. The standard equation of a circle with a center at the origin \((0,0)\) is \(x^2 + y^2 = r^2\), where \(r\) represents the radius of the circle.

Circles can have various positions on the coordinate plane. If the center is at \( (h, k) \), the equation changes to \( (x - h)^2 + (y - k)^2 = r^2 \). This shift allows you to place a circle anywhere, while still maintaining its perfect roundness and symmetry.

It's also important to recognize the implications of constants in an equation. In \(x^2 + y^2 = 5\), the radius is \(\sqrt{5}\), emphasizing that while a circle's format is consistent, its size and placement on the plane can vary widely, altering its interaction with other geometric elements, like tangents.

Tangent to a Circle

A tangent to a circle is a straight line that touches the circle at exactly one point. This unique point is called the point of tangency. A characteristic property of a tangent line is that it is perpendicular to the radius at the point of tangency.

Understanding how to find the equation of a tangent to a circle is crucial when dealing with circles intersecting with lines. For a circle centered at the origin \(x^2 + y^2 = r^2\), the equation of the tangent at any point \(x_1, y_1\) on the circle can be expressed as \(xx_1 + yy_1 = r^2\). Here, \(x_1, y_1\) are coordinates on the circle itself, ensuring the tangent meets the circle only at this specific point.

Finding this line is critical in problems where the tangent interacts with another circle, either intersecting or just grazing it, thereby forming the basis for further calculations, such as finding the exact points of contact.

Point of Contact

The point of contact is the exact location where a tangent touches a circle. This point is incredibly important as it confirms the tangent's validity and position relative to the circle.

To find this point, you generally solve for a coordinate that satisfies both the circle's equation and the tangent's equation simultaneously. When you draw both the circle and its tangent, the point where the two touch is the point of contact \(x_1, y_1\).

In our exercise, the point of contact is found by solving the system of equations set by the circle and tangent. Each solution represents a possible point where the line meets the circle. It is essential to correctly identify this point to understand both the circle's tangent relationship and each circle involved in more complex geometric arrangements.

Equation of a Circle

The equation of a circle provides you with a precise mathematical representation of the circle in a coordinate plane. This equation, generally expressed as \(x^2 + y^2 + 2gx + 2fy + c = 0\), gives valuable insights into the circle's key features.

By transforming and solving this form, you can easily determine the circle's center and radius. Completing the square will allow you to convert the general form into the center-radius form \((x - h)^2 + (y - k)^2 = r^2\), where \(h, k\) are the coordinates of the center, and \(r\) is the radius.

In the given exercise, two circles are compared. The first is straightforward \((x^2 + y^2 = 5\), indicating a simple circle centered at the origin. The second equation \((x^2 + y^2 - 8x + 6y + 20 = 0\)") is modified to identify another circle and examine its interaction with the tangent. Analyzing these equations reveals much about the positions and dimensions of each circle, crucial for solving complex geometric problems.