JEE Main Coordinate Geometry: Circles – Chapter Test

Topics in this test

Difficulty: 8 medium · 12 hard

Circles MCQ questions with answers

The first 3 of 20 questions. Unlock the series for the rest, each with a solution.

1. Q1. For what values of k does the equation x² + y² − 2x + 4y + k = 0 represent a real circle?

   k के किन मानों के लिए समीकरण x² + y² − 2x + 4y + k = 0 एक वास्तविक वृत्त को निरूपित करता है?

   • A. k > 5
   • B. k = 5
   • C. k < 5
   • D. k ≤ 5
   Solution

   The radius is √(g² + f² − c) = √(1 + 4 − k) = √(5 − k), which is real and positive only when 5 − k > 0, i.e. k < 5.

   त्रिज्या √(g² + f² − c) = √(1 + 4 − k) = √(5 − k) है, जो केवल तभी वास्तविक और धनात्मक होती है जब 5 − k > 0, अर्थात् k < 5।

2. Q2. The equation of the normal to the circle x² + y² = 25 at the point (3, 4) is:

   बिंदु (3, 4) पर वृत्त x² + y² = 25 के अभिलंब (normal) का समीकरण है:

   • A. 4x + 3y = 0
   • B. 3x + 4y = 25
   • C. 3x − 4y = 0
   • D. 4x − 3y = 0
   Solution

   A normal to a circle passes through its centre, so it joins the origin and (3, 4), giving the line y = (4/3)x, i.e. 4x − 3y = 0.

   वृत्त का अभिलंब उसके केंद्र से होकर जाता है, अतः यह मूल बिंदु और (3, 4) को मिलाता है, जिससे रेखा y = (4/3)x, अर्थात् 4x − 3y = 0 प्राप्त होती है।

3. Q3. Arrange the points P(1, 1), S(3, 3), Q(4, 2) and R(0, 5) in increasing order of their power with respect to the circle x² + y² = 16.

   बिंदुओं P(1, 1), S(3, 3), Q(4, 2) तथा R(0, 5) को वृत्त x² + y² = 16 के सापेक्ष उनकी शक्ति के बढ़ते क्रम में व्यवस्थित कीजिए।

   • A. P, S, R, Q
   • B. P, Q, S, R
   • C. S, P, Q, R
   • D. P, S, Q, R
   Solution

   The power of (x₁, y₁) is x₁² + y₁² − 16, giving −14 for P, 2 for S, 4 for Q and 9 for R, so the increasing order is P, S, Q, R.

   बिंदु (x₁, y₁) की शक्ति x₁² + y₁² − 16 होती है, जिससे P के लिए −14, S के लिए 2, Q के लिए 4 तथा R के लिए 9 प्राप्त होता है, अतः बढ़ता क्रम P, S, Q, R है।

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