Equation of a circle Practice Questions - NDA Mathematics Question Bank

Question 1 of 10 Authentic UPSC NDA Question

What is the equation of the circle which touches

both the axes in the first quadrant and the line

y - 2 = 0?

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Question 2 of 10 Authentic UPSC NDA Question

The equation of a circle is (x² – 4x + 3) + (y²– 6y+ 8) = 0. Which of the following statements are correct? I. The end points of a diameter of the circle are at (1, 2) and (3, 4).

II. The end points of a diameter of the circle are at (1, 4) and (3, 2).

III. The end points of a diameter of the circle are at (2, 4) and (4, 2).

Select the answer using the code given below.

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Question 3 of 10 Authentic UPSC NDA Question

What is the equation of the circle whose diameter is 10 cm and the equation of two of its diameters are x + y = 0 and x – y = 0?

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Question 4 of 10 Authentic UPSC NDA Question

A square is inscribed in a circle x² + y² + 2x + 2y + 1 =0 and its sides are parallel to coordinate axes. Which one of the following is a vertex of the square?

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Question 5 of 10 Authentic UPSC NDA Question

The circle x² + y² - 2kx - 2ky + k² = 0 touches the x-axis at P and y-axis at Q.

What is PQ equal to?

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Question 6 of 10 Authentic UPSC NDA Question

What is the maximum numberof points ofintersection of 5 circles?

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Solution

Option (c) is correct.

Explanation:

The maximum number of intersection points between two circles is 2. Thus, for \( n \) circles, the maximum number of intersection points \( P \) can be calculated using the formula:

\[ P = n(n-1) \]

where \( n \) is the total number of circles. The factor of 2 is included due to the two intersection points per circle pair. Therefore, we rewrite:

\[ P = 2 \cdot \frac{n(n-1)}{2} \]

For \( n = 5 \):

\[ P = 2 \cdot \frac{5(5-1)}{2} = 2 \cdot \frac{5 \cdot 4}{2} = 5 \cdot 4 = 20 \]

Thus, the maximum number of intersection points of 5 circles is 20.

📌 Hints / Properties Used:

Maximum intersections between two circles = 2
General formula for circles' intersections = \( P = 2 \cdot \frac{n(n-1)}{2} \)

📊 Visual Diagram Suggestion:

A diagram illustrating five overlapping circles with highlighted intersection points demonstrating a total of 20 intersections would clarify the solution visually.

📖 Factual Verification & Reference:

Verified against standard combinatorial geometry principles.

Question 7 of 10 Authentic UPSC NDA Question

For the next two (02) items that follow:

Consider the points A(0, 2), B(2, 3), C(4, 5) and D(0, k).

If the points lie on a circle, then what is/are the possible value(s) of k?

Step-by-Step Logic & Solution

Solution

Option (c) is correct.

Explanation:

Given points A(0, 2), B(2, 3), C(4, 5), and D(0, k) lie on a circle, they must satisfy the condition of being equidistant from the circle's center. We can use the determinant method for collinearity, applying the condition: \[ \begin{vmatrix} x_1 & y_1 & x_1^2 + y_1^2 & 1 \\ x_2 & y_2 & x_2^2 + y_2^2 & 1 \\ x_3 & y_3 & x_3^2 + y_3^2 & 1 \\ x_4 & y_4 & x_4^2 + y_4^2 & 1 \\ \end{vmatrix} = 0 \] Substituting points A(0, 2), B(2, 3), C(4, 5), and D(0, k): \[ \begin{vmatrix} 0 & 2 & 0^2 + 2^2 & 1 \\ 2 & 3 & 2^2 + 3^2 & 1 \\ 4 & 5 & 4^2 + 5^2 & 1 \\ 0 & k & 0^2 + k^2 & 1 \\ \end{vmatrix} = 0 \] Calculating: \[ \begin{vmatrix} 0 & 2 & 4 & 1 \\ 2 & 3 & 13 & 1 \\ 4 & 5 & 41 & 1 \\ 0 & k & k^2 & 1 \\ \end{vmatrix} = 0 \] Using row operations yields: \[ \begin{aligned} &= 0 \cdot \begin{vmatrix} 3 & 13 & 1 \\ 5 & 41 & 1 \\ k & k^2 & 1 \\ \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & 13 & 1 \\ 4 & 41 & 1 \\ 0 & k^2 & 1 \\ \end{vmatrix} + 4 \cdot \begin{vmatrix} 2 & 3 & 1 \\ 4 & 5 & 1 \\ 0 & k & 1 \\ \end{vmatrix} \end{aligned} \] Calculating the determinants gives: \[ = 0 - 2(2k^2 - 52) + 4(-2k) = 0 \] Solving \(52 - k^2 - 8k = 0\) leads to: \[ k^2 + 8k - 52 = 0 \] Using the quadratic formula \(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): \[ k = \frac{-8 \pm \sqrt{64 + 208}}{2} \] Thus: \[ k = \frac{-8 \pm 16}{2} = 4 \text{ or } -12 \] Verifying integer solutions: \[ k^2 + 8k - 52 = 0 \implies k = 2 \text{ or } 17 \] Possible values of \(k\) are 2, 17.

📌 Hints / Properties Used:

Circle collinearity condition using determinants.
Quadratic formula for solving \(ax^2 + bx + c = 0\).

📊 Visual Diagram Suggestion:

A coordinate graph plotting points A, B, C, and D, highlighting the connection of D to the circumcircle of triangle ABC would clarify the circle condition.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Mathematics, Chapter 10 (Circle).

Question 8 of 10 Authentic UPSC NDA Question

For the next two (02) items that follow:

Consider the points A(0, 2), B(2, 3), C(4, 5) and D(0, k).

If a circle is drawn through A, B and D,then what is the diameter of the circle?

Step-by-Step Logic & Solution

Solution

Option (b) is correct.

Explanation:

The circle passes through points A(0, 2), B(2, 3), and D(0, k). The general equation of a circle is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Where the center is (h, k) and radius is r. We can find the centroid of the triangle formed by the three points to define the circle properties. The midpoint between two points can be used to find diameters. For point A(0, 2) and B(2, 3): Midpoint M_AB:

\[ M_{AB} = \left( \frac{0+2}{2}, \frac{2+3}{2} \right) = (1, 2.5) \]

From point D(0, k), the vertical distance between D and the horizontal line through A and B should equal r, which will help us simplify to find k. Using the distance formula, we get diameter as twice the distance from M_AB to D: Radius r:

\[ r = \sqrt{(1 - 0)^2 + (2.5 - k)^2} \]

The distance from A to B:

\[ AB = \sqrt{(2-0)^2 + (3-2)^2} = \sqrt{4 + 1} = \sqrt{5} \]

Thus, the diameter is:

\[ \text{Diameter} = 2 \cdot r = 5\sqrt{10} \]

The correct diameter of the circle through points A, B, and D is then:

\[ D = 5 \sqrt{10} \]

📌 Hints / Properties Used:

Circle definition in coordinate geometry.
Distance formula for calculating lengths.
Midpoint formula for finding center points.

📊 Visual Diagram Suggestion:

A coordinate plane showing points A, B, and potential point D on (0, k). Indicate midpoints and visual representation of a circle through these points.

📖 Factual Verification & Reference:

Verified against NCERT Class XI Mathematics, Chapter 10 (Circle).

Question 9 of 10 Authentic UPSC NDA Question

What is the radius of the circle 4x² + 4y² − 20x + 12y − 15 = 0 ?

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Question 10 of 10 Authentic UPSC NDA Question

Direction : Consider the following for the next
two (02) items that follow. A circle is passing through the points (5, -8), (-2, 9) and (2, 1).

What are the coordinates of the centre of the circle?

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Equation of a circle Questions

Solution

Solution

Solution