Problem 87 - A Complete Resource Book in Mathematics for JEE Main

Question

A tangent drawn from the point \((4,0)\) to the circle \(x^{2}+\) \(y^{2}=8\) touches it at a point \(A\) in the first quadrant. The coordinates of another point \(B\) on the circle such that \(A B=4\), are (A) \((2,-2)\) (B) \((-2,2)\) (C) \((2,2)\) (D) \((-2,-2)\)

Step-by-Step Solution

Answer

The coordinates of point B are (2,-2).

Step 1: Find the equation of the tangent

Recall that the equation of the circle is \(x^2 + y^2 = 8\). The tangent from a point \((x_1, y_1)\) to a circle \(x^2 + y^2 = r^2\) has the equation \(xx_1 + yy_1 = r^2\). For the point \((4,0)\), the tangent equation is \(4x = 8\), which simplifies to \(x = 2\).

Step 2: Determine Point A

Since the tangent line \(x = 2\) meets the circle in the first quadrant, we substitute \(x = 2\) into the circle's equation: \((2)^2 + y^2 = 8\). Solving for \(y\), we get \(y^2 = 4\), thus \(y = 2\) (since \(y > 0\) for the first quadrant). Hence, the coordinates of point \(A\) are \((2, 2)\).

Step 3: Use Distance Formula for Point B

The distance \(AB\) is given as 4. We have \(A(2, 2)\), let's consider \(B(x, y)\) on the circle. The distance between \(A\) and \(B\) is given by the formula \(\sqrt{(x - 2)^2 + (y - 2)^2} = 4\).

Step 4: Solve for Coordinates of Point B

Substitute \(x^2 + y^2 = 8\) (since \(B\) lies on the circle) in the distance formula found in step 3: \[(x - 2)^2 + (y - 2)^2 = 16\]Expanding this gives:\[x^2 - 4x + 4 + y^2 - 4y + 4 = 16\]Since \(x^2 + y^2 = 8\):\[8 - 4x - 4y + 8 = 16\]Which simplifies to:\[-4x - 4y = 0\]Therefore, \(x + y = 0\). Since \(B\) is on the circle, try the options Substitute \(x = 2\) into \(x+y = 0\), then \(y = -2\).Thus, \((x,y) = (2,-2)\) satisfies both the circle's equation \(x^2 + y^2 = 8\) and \((x - 2)^2 + (y - 2)^2 = 16\).

Step 5: Verify Point B

Check option \(B(x=2, y=-2)\). => For \(B(2, -2)\):\(2^2 + (-2)^2 = 8\) (Circle's equation holds)Calculate distance \(AB\) again: => \(\sqrt{(2 - 2)^2 + (-2 - 2)^2} = \sqrt{0 + 16} = 4\).It satisfies both conditions.

Key Concepts

Circle Equation

The circle equation is a mathematical representation of a circle on a coordinate plane. It is centered at the origin with a standard form given by \(x^2 + y^2 = r^2\), where \(r\) is the radius of the circle. In the provided exercise, the circle given by the equation \(x^2 + y^2 = 8\) suggests that its center is at the origin \((0,0)\) and has a radius of \(\sqrt{8} = 2\sqrt{2}\).

When solving problems involving circles, it is important to remember:

• The circle's equation can help identify points that lie on the circle.
• Any line, such as a tangent or a radius, is related mathematically to the circle's equation.
• Substitution can be used to test whether a particular point falls within the circle.

Understanding how to manipulate and interpret this equation is crucial for solving problems in coordinate geometry.

Tangent Line

A tangent line to a circle is a straight line which touches the circle at exactly one point. This point is referred to as the point of tangency. In the exercise, the tangent line from point \((4,0)\) to the circle \(x^2 + y^2 = 8\) touches the circle at point \(A\) in the first quadrant.

To find the equation of the tangent line, we can use the formula \(xx_1 + yy_1 = r^2\), where \((x_1, y_1)\) is a point from which the tangent is drawn, and \(r\) is the radius of the circle. When applied to point \((4,0)\), it simplifies to \(4x = 8\), further reduced to \(x = 2\). This indicates that the tangent is a vertical line intersecting the x-axis at \(x = 2\).

Key points to remember about tangent lines are:

• They only touch the circle at one point and do not cross through it.
• The radius drawn to the point of tangency is perpendicular to the tangent line.
• Using the tangent line equation helps find the specific point of contact on the circle.

Distance Formula

The distance formula is a crucial tool for calculating the distance between two points in a plane. It is given by: \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). In this exercise, it helps find the distance between points \(A\) and \(B\) on the circle.

We know the coordinates of point \(A\) is \((2, 2)\) from solving the tangent problem, and the distance \(AB\) is known to be 4. To determine point \(B(x, y)\), we set up: \(\sqrt{(x - 2)^2 + (y - 2)^2} = 4\). Ensuring this equation holds while also considering the circle's equation \(x^2 + y^2 = 8\) allows us to find the valid position of \(B\).

Key takeaways for using the distance formula include:

• Ensuring the points being compared align with known conditions (e.g. one lies on a circle).
• Solving the equation can result in multiple solutions; context helps choose the correct one.

Mastery of this formula aids in solving geometric problems involving distances.

Quadrants

In coordinate geometry, the coordinate plane is divided into four quadrants by the x-axis and y-axis. Each quadrant has specific characteristics regarding the sign of x and y values. The point \(A\), found in the exercise as \((2, 2)\), lies in the first quadrant, which is characterized by both x and y being positive.

Here's a simple breakdown of the quadrants:

• First Quadrant - x > 0, y > 0
• Second Quadrant - x < 0, y > 0
• Third Quadrant - x < 0, y < 0
• Fourth Quadrant - x > 0, y < 0

Understanding the properties of quadrants is key when identifying or predicting the position of a point within the plane, allowing for informed decision-making when solving geometrical problems.