Problem 56 - A Complete Resource Book in Mathematics for JEE Main

Question

A circle touches the line \(y=x\) at \(a\) point \(P\) such that \(O P=4 \sqrt{2}\), where \(O\) is the origin. The circle contains the point \((-10,2)\) in its interior and the length of its chord on the line \(x+y=0\) is \(6 \sqrt{2}\). The equation of the circle is (A) \(x^{2}+y^{2}+18 x-2 y+32=0\) (B) \(x^{2}+y^{2}-18 x-2 y+32=0\) (C) \(x^{2}+y^{2}+18 x+2 y+32=0\) (D) none of these

Step-by-Step Solution

Answer

The equation of the circle is (D) none of these.

Step 1: Identify the Perpendicular Condition

The circle touches the line \(y=x\) at a point \(P\). Therefore, the center of the circle \((h,k)\) must satisfy the condition that the line \(y=x\) is perpendicular to the radius of the circle at point \(P\). Thus, the slope of OP should equal the slope of the perpendicular radius. Hence, \(h=k\).

Step 2: Use the Distance Formula

The distance \(OP\) from the origin \((0,0)\) to the point \(P(h,h)\) is given as \(4\sqrt{2}\). Using the distance formula, we have: \(\sqrt{h^2 + h^2} = 4\sqrt{2}\). Solve for \(h\): \(\sqrt{2h^2} = 4\sqrt{2}\). This simplifies to \(h=\pm4\).

Step 3: Choose the Appropriate Center of the Circle

We determine that the circle is centered at \((4,4)\) because it must contain the point \((-10,2)\) inside it, and \((-4, -4)\) would be farther away, potentially not satisfying the interior condition.

Step 4: Determine the Radius Using Point Inside the Circle

The circle contains the point \((-10,2)\) inside it, so calculate the distance from the center \((4,4)\) to \((-10,2)\): \(\sqrt{(4+10)^2 + (4-2)^2} = \sqrt{196 + 4} = 14\). Thus, the radius \(r\) of the circle is greater than \(14\).

Step 5: Analyze the Chord Condition

The length of the chord on the line \(x+y=0\) is \(6\sqrt{2}\). The perpendicular distance from the center \((4,4)\) to the line \(x+y=0\) can be found using the formula \(|h+k+c|/\sqrt{a^2+b^2}\), which gives \(|4+4+0|/\sqrt{2} = 8/\sqrt{2} = 4\sqrt{2}\). Since the length of the chord is \(2\sqrt{r^2 - d^2}\), solve \(6\sqrt{2} = 2\sqrt{r^2 - 32}\), leading to \(r^2 = 50\). Hence, \(r = \sqrt{50} = 5\sqrt{2}\).

Step 6: Write the Equation of the Circle

With center \((4,4)\) and radius \(r = 5\sqrt{2}\), the equation of the circle is: \[(x-4)^2 + (y-4)^2 = (5\sqrt{2})^2 = 50\]Expand to get: \[(x^2 - 8x + 16) + (y^2 - 8y + 16) = 50\]Combine terms: \[x^2 + y^2 - 8x - 8y + 32 = 50\]Simplify: \[x^2 + y^2 - 8x - 8y + 32 - 50 = 0\]Final form: \[x^2 + y^2 - 8x - 8y - 18 = 0\]. This matches with none of the given options, meaning the answer is (D) none of these.

Key Concepts

Circle Touches Line

When we say a circle touches a line, it means the circle is tangent to the line. This means that the line only touches the circle at exactly one point. At this touching point, the line is perpendicular to the radius of the circle.
The given problem states that the circle touches the line \(y = x\) at a point \(P\). This condition implies that the center of the circle lies on a line perpendicular to \(y = x\). For the line \(y = x\), it means that the slope of a line perpendicular to it is \(-1\).
Since the circle's center must also align with this slope, it satisfies the condition that the center \((h, k)\) must lie such that \(h = k\), meaning the center is equidistant from the x and y-axis. The fact that the specific distance from the origin to this center is \(4\sqrt{2}\) helps pinpoint the circle's center more accurately.

Chord Length

A chord is a line segment with both endpoints lying on the circle. A key relationship exists between the length of a chord, the distance from the center of the circle to the chord, and the radius of the circle.
In this problem, a chord of the circle is on the line \(x + y = 0\), and its length is \(6\sqrt{2}\). The center \((4, 4)\) and the distance from this center to the line is used, given by the formula:

• \(|ax_1 + by_1 + c| / \sqrt{a^2 + b^2}\) gives the distance from the center to the line.
• Here, \(a = 1, b = 1, c = 0\), and the point \((h, k)\) is \((4, 4)\).

The distance calculated becomes \(4\sqrt{2}\) and relates directly to solving for the radius using the chord length condition, \(6\sqrt{2} = 2\sqrt{r^2 - d^2}\). Solving it gives \(r = 5\sqrt{2}\). This calculation ensures that the circle encompasses the chord correctly, showing the intrinsic geometry of circle-line interactions.

Point Inside Circle

To say a point is inside a circle means that the distance from the center of the circle to this point is less than the radius of the circle. This relationship is vital in determining if a given point lies within, on, or outside the circle.
For our exercise, the point \((-10, 2)\) is said to be inside the circle. Therefore, we find the straight-line distance from the circle's center \((4, 4)\) to this point using the Pythagorean theorem:

• Distance = \(\sqrt{(4 + 10)^2 + (4 - 2)^2} = 14\).

This tells us that the radius of the circle must be greater than \(14\) since the point lies inside. The condition of containing this point affects the decision to choose the correct center \((4, 4)\), as any other center might not satisfy this condition. This foundational understanding of circle geometry ensures that the equations of the circle correctly account for all listed conditions.