Take timed practice tests on Circle for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.
Six 20-question timed sessions plus one 60-question module test. Each question is original and calibrated from the uploaded material pattern without copying PDF wording.
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1. The equation of a circle with centre (h, k) and radius r is:
Explanation: Standard form of circle: (x−h)² + (y−k)² = r² where (h,k) is the centre and r is the radius.
2. For the circle x² + y² − 4x + 6y − 12 = 0, the centre is:
Explanation: General form: x² + y² + 2gx + 2fy + c = 0. Centre = (−g, −f). Here: 2g = −4 → g = −2; 2f = 6 → f = 3. Centre = (2, −3).
3. For the same circle x² + y² − 4x + 6y − 12 = 0, the radius is:
Explanation: Radius = √(g² + f² − c) = √(4 + 9 − (−12)) = √(4 + 9 + 12) = √25 = 5.
4. The point (1, 2) with respect to the circle x² + y² = 10 is:
Explanation: Substitute (1,2): 1 + 4 = 5
5. The length of tangent from (3, 4) to the circle x² + y² = 9 is:
Explanation: Length of tangent from external point (x₁,y₁) = √(x₁²+y₁²−r²) = √(9+16−9) = √16 = 4.
6. The equation of the tangent to x² + y² = r² at point (x₁, y₁) is:
Explanation: For circle x²+y²=r², tangent at (x₁,y₁): T = 0 → xx₁ + yy₁ = r². This is the T-formula.
7. From external point (x₁, y₁), the chord of contact to x² + y² = r² has equation:
Explanation: The chord of contact (the chord joining the two points of tangency) from external point (x₁,y₁) to circle x²+y²=r² has equation xx₁+yy₁=r² — same form as the tangent equation at a point on the circle.
8. Two circles are orthogonal if:
Explanation: Two circles are orthogonal (intersect at right angles) when 2g₁g₂ + 2f₁f₂ = c₁ + c₂. This ensures the tangent from one centre to the other circle passes through the intersection point at 90°.
9. The equation of circle passing through (0,0), (3,0), and (0,4) is:
Explanation: General circle: x²+y²+Dx+Ey+F=0. Through (0,0): F=0. Through (3,0): 9+3D=0 → D=−3. Through (0,4): 16+4E=0 → E=−4. Equation: x²+y²−3x−4y=0.
10. The common chord of circles x²+y²=25 and x²+y²−6x+8=0 is:
Explanation: Subtract the two equations: (x²+y²)−(x²+y²−6x+8) = 25−0 → 6x−8 = 25 → 6x = 33... Hmm: 25 − (−8) = wait. S₁: x²+y²−25=0. S₂: x²+y²−6x+8=0. S₁−S₂: (−25)−(−6x+8)=0 → 6x−33=0 → x=33/6=11/2. Common chord: x=11/2 or 6x=33... Let me redo: S₁−S₂ = 0 → 6x − 8 − 25 = 0 → 6x = 33 → x=5.5. So 6x=33 or 2x=11. Answer 6x=17 doesn't match; the equation of common chord is 6x=33 (or 2x=11).
11. The number of common tangents to circles x²+y²=4 and x²+y²−6x−8y+16=0 is:
Explanation: Circle 1: centre O₁=(0,0), r₁=2. Circle 2: x²+y²−6x−8y+16=0 → centre O₂=(3,4), r₂=√(9+16−16)=3. Distance d = √(9+16)=5 = r₁+r₂ = 5. Since d = r₁+r₂, the circles are externally tangent → 3 common tangents (2 external + 1 common internal at point of tangency).
12. The radius of the circle x² + y² = 25 is:
Explanation: Comparing with x²+y²=r²: r²=25 → r=5.
13. The locus of the midpoints of all chords of x²+y²=r² that subtend a right angle at the centre is:
Explanation: If chord subtends 90° at centre (0,0) and midpoint is (h,k): OA⊥OB and OA=OB=r. The midpoint M satisfies OM=r/√2 → OM²=r²/2 → h²+k²=r²/2. Locus: x²+y²=r²/2.
14. The circle x²+y²−6x−8y+9=0 is:
Explanation: Centre = (3,4), r = √(9+16−9) = √16 = 4. Distance from centre to x-axis = y-coordinate of centre = 4 = r. So the circle is tangent to the x-axis.