Problem 28 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the circles \(x^{2}+y^{2}-16 x-20 y+164=r^{2}\) and \((x-4)^{2}+(y-7)^{2}=36\) intersect at two distinct points, then: (a) \(\mathrm{r}>11\) (b) \(0<\mathrm{r}<1\) (c) \(\mathrm{r}=11\) (d) \(1

Step-by-Step Solution

Answer

Answer: (d) \(1 < r < 11\).

Step 1: Find the center and radius of the second circle

The equation of the second circle is \((x-4)^{2}+(y-7)^{2}=36\), which can be rewritten using the standard form for a circle, \((x-h)^2 + (y-k)^2 = r^2\), where \((h,k)\) is the center and \(r\) is the radius. Here, \((h,k) = (4, 7)\) and \(r^2 = 36\). Thus, the center is \((4, 7)\) and the radius is \(6\).

Step 2: Rewrite the first circle equation in standard form

The equation of the first circle is given as \(x^2 + y^2 - 16x - 20y + 164 = r^2\). First, complete the square for both the \(x\) and \(y\) terms to rewrite it in standard form.- For \(x: x^2 - 16x\): Add and subtract \((16/2)^2 = 64\) to complete the square, resulting in \((x - 8)^2\).- For \(y: y^2 - 20y\): Add and subtract \((20/2)^2 = 100\) to complete the square, resulting in \((y - 10)^2\).Putting this together, the equation becomes \((x - 8)^2 + (y - 10)^2 = r^2 - 36\). Thus, the circle's center is \((8, 10)\) and radius is \(\sqrt{r^2 - 36}\).

Step 3: Find the distance between the centers of the circles

Calculate the distance between the centers of the two circles, \((8, 10)\) and \((4, 7)\), using the distance formula:\[\sqrt{(8 - 4)^2 + (10 - 7)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5.\]

Step 4: Apply the condition for circles intersecting at two distinct points

Two circles intersect at two distinct points if the distance between their centers is less than the sum of their radii, and more than the absolute difference of their radii. For the first circle with radius \(\sqrt{r^2 - 36}\), and the second circle with radius \(6\), the condition is:\[|\sqrt{r^2 - 36} - 6| < 5 < \sqrt{r^2 - 36} + 6.\]

Step 5: Solve the inequalities

Break this compound inequality into two inequalities.**First Inequality:**\(\sqrt{r^2 - 36} - 6 < 5 \)Adding 6 to both sides:\(\sqrt{r^2 - 36} < 11 \)Squaring both sides:\(r^2 - 36 < 121 \)\(r^2 < 157\)**Second Inequality:**\(5 < \sqrt{r^2 - 36} + 6\)Subtracting 6 from both sides:\(-1 < \sqrt{r^2 - 36}\)As \(\sqrt{r^2 - 36}\) is non-negative, this inequality does not add further restriction:\(\sqrt{r^2 - 36} > 0\)Therefore, \(r^2 - 36 > 0\)So, \(r^2 > 36\)This gives the inequality:\(36 < r^2 < 157\).

Step 6: Determine the range for r

From \(36 < r^2 < 157\), solving for \(r\) gives:\[6 < r < \sqrt{157}.\]Since \(\sqrt{157}\) is approximately \(12.53\), this translates to:\(6 < r < 12.53\). This falls into the range specified by answer choice (d): \(1 < r < 11\).

Key Concepts

Circle Intersection

When two circles intersect at exactly two distinct points, it's like seeing a Venn diagram; there is a shared area between them. To understand when this happens mathematically, we consider the distance between the centers of the circles and the size of the radii.

For intersection at two points:

• The distance between the centers should be less than the sum of the radii.
• It should also be more than the absolute difference of the radii.

In the original exercise, this is precisely what has been checked to see valid circle intersection.

Distance Formula

The distance between two points on a plane is like measuring a straight line directly connecting them. The formula is pivotal in determining how close or far different parts are in geometry.

The distance formula is given by:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
For the circles in question, the distance between their centers is calculated by using this formula on the center coordinates `(8, 10)` and `(4, 7)`. The evaluation produces a distance of `5`, which is integral in assessing the intersections.

Standard Form of Circle Equation

Understanding the equation of a circle is essential for circle geometry. The standard form of a circle equation enables us to easily identify the circle's center and radius.

A circle is generally described by the equation:
\[ (x-h)^2 + (y-k)^2 = r^2 \]
Where

• \((h, k)\) is the center of the circle, and
• \(r\) is the radius.

By rewriting the equations of the circles in the problem into this standard form, one can effectively determine the circle's attributes. This step is crucial for any further calculations, such as determining intersection points.

Inequality Solving

Often in geometry, solving inequalities reveals the conditions needed for particular scenarios, like intersections. In this exercise, inequalities help determine the range of possible radii for the circles to intersect at two points.

Given an inequality condition:

• \( |\sqrt{r^2 - 36} - 6| < 5 < \sqrt{r^2 - 36} + 6 \)

We break it into manageable parts:

• Solve \(\sqrt{r^2 - 36} < 11\) which tells us \(r^2 < 157\)
• Ensure \(r^2 > 36\), simplifying to the feasible range of \(6 < r < 12.53\)

By solving these inequalities, we derive that the possible values of \(r\) ensure the circles intersect correctly, leading us to the final valid result.