Find the equation of the circle with centre (0,2) and radius 2.
Hint. Substitute h=0, k=2, r=2 directly into (x-h)^2+(y-k)^2=r^2.
Here h=0, k=2, r=2. Substituting: (x-0)^2+(y-2)^2=2^2.
✦ Working through each part gives: x^2+(y-2)^2=4.
Find the equation of the circle with centre (-2,3) and radius 4.
Hint. Substitute h=-2, k=3, r=4 into (x-h)^2+(y-k)^2=r^2.
Here h=-2, k=3, r=4. Substituting: (x+2)^2+(y-3)^2=16.
✦ Working through each part gives: (x+2)^2+(y-3)^2=16.
Find the equation of the circle with centre (1/2, 1/4) and radius 1/12.
Hint. Substitute h=1/2, k=1/4, r=1/12 into (x-h)^2+(y-k)^2=r^2. Confirmed against the official page image, since the raw PDF text extraction had reversed the coordinate order.
Here h=1/2, k=1/4, r=1/12. Substituting: (x-1/2)^2+(y-1/4)^2=(1/12)^2=1/144.
✦ Working through each part gives: (x-1/2)^2+(y-1/4)^2=1/144.
Find the equation of the circle with centre (1,1) and radius sqrt(2).
Hint. Substitute h=1, k=1, r=sqrt(2) into (x-h)^2+(y-k)^2=r^2.
Here h=1, k=1, r=sqrt(2). Substituting: (x-1)^2+(y-1)^2=2.
✦ Working through each part gives: (x-1)^2+(y-1)^2=2.
Find the equation of the circle with centre (-a,-b) and radius sqrt(a^2-b^2).
Hint. Substitute h=-a, k=-b, r=sqrt(a^2-b^2) into (x-h)^2+(y-k)^2=r^2.
Here h=-a, k=-b, r=sqrt(a^2-b^2). Substituting: (x+a)^2+(y+b)^2=a^2-b^2.
✦ Working through each part gives: (x+a)^2+(y+b)^2=a^2-b^2.
Find the centre and radius of the circle (x+5)^2+(y-3)^2=36.
Hint. Compare directly with (x-h)^2+(y-k)^2=r^2 — no expansion needed since this is already in standard form.
Comparing (x+5)^2+(y-3)^2=36 with (x-h)^2+(y-k)^2=r^2 gives h=-5, k=3, r^2=36 so r=6.
✦ Working through each part gives: Centre (-5, 3), radius 6.
Find the centre and radius of the circle x^2+y^2-4x-8y-45=0.
Hint. Group the x-terms and y-terms separately, then complete the square on each.
Grouping: (x^2-4x)+(y^2-8y)=45. Completing the square: (x^2-4x+4)+(y^2-8y+16)=45+4+16=65, i.e. (x-2)^2+(y-4)^2=65. Comparing with (x-h)^2+(y-k)^2=r^2 gives centre (2,4) and radius sqrt(65).
✦ Working through each part gives: Centre (2, 4), radius sqrt(65).
Find the centre and radius of the circle x^2+y^2-8x+10y-12=0.
Hint. Group the x-terms and y-terms separately, then complete the square on each.
Grouping: (x^2-8x)+(y^2+10y)=12. Completing the square: (x^2-8x+16)+(y^2+10y+25)=12+16+25=53, i.e. (x-4)^2+(y+5)^2=53. Comparing with (x-h)^2+(y-k)^2=r^2 gives centre (4,-5) and radius sqrt(53).
✦ Working through each part gives: Centre (4, -5), radius sqrt(53).
Find the centre and radius of the circle 2x^2+2y^2-x=0.
Hint. First divide through by 2 to make the x^2 and y^2 coefficients 1, then complete the square.
Dividing by 2: x^2+y^2-x/2=0. Completing the square on x: (x^2-x/2+1/16)+y^2=1/16, i.e. (x-1/4)^2+y^2=1/16. Comparing with (x-h)^2+(y-k)^2=r^2 gives centre (1/4, 0) and radius sqrt(1/16)=1/4.
✦ Working through each part gives: Centre (1/4, 0), radius 1/4.
Where students slip. Forgetting to divide the whole equation by the leading coefficient (2) before completing the square gives a centre and radius scaled incorrectly.
Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x+y=16.
Hint. Use the general form x^2+y^2+2gx+2fy+c=0. Substituting both points gives two equations in g, f, c; the centre condition (-g,-f) on 4x+y=16 gives a third.
Substituting (4,1): 17+8g+2f+c=0 ...(i). Substituting (6,5): 61+12g+10f+c=0 ...(ii). Centre (-g,-f) lies on 4x+y=16: -4g-f=16, i.e. f=-4g-16 ...(iii). Subtracting (i) from (ii): 44+4g+8f=0, i.e. g+2f=-11 ...(iv). Substituting (iii) into (iv): g+2(-4g-16)=-11, so g-8g-32=-11, giving -7g=21, g=-3. Then f=-4(-3)-16=-4. From (i): 17+8(-3)+2(-4)+c=0, so 17-24-8+c=0, giving c=15.
✦ Working through each part gives: x^2+y^2-6x-8y+15=0.
Find the equation of the circle passing through the points (2,3) and (-1,1) and whose centre is on the line x-3y-11=0.
Hint. Use the general form x^2+y^2+2gx+2fy+c=0, substitute both points, and use the centre condition to solve for g, f, c.
Substituting (2,3): 13+4g+6f+c=0 ...(i). Substituting (-1,1): 2-2g+2f+c=0 ...(ii). Centre (-g,-f) lies on x-3y-11=0: -g+3f-11=0, i.e. g=3f-11 ...(iii). Subtracting (ii) from (i): 11+6g+4f=0 ...(iv). Substituting (iii) into (iv): 11+6(3f-11)+4f=0, so 11+18f-66+4f=0, giving 22f=55, f=5/2. Then g=3(5/2)-11=-7/2. From (ii): 2-2(-7/2)+2(5/2)+c=0, so 2+7+5+c=0, giving c=-14.
✦ Working through each part gives: x^2+y^2-7x+5y-14=0.
Find the equation of the circle with radius 5 whose centre lies on the x-axis and passes through the point (2,3).
Hint. A centre on the x-axis has the form (h,0). Set the distance from (h,0) to (2,3) equal to 5 and solve for h.
Let the centre be (h,0). Since the radius is 5: (2-h)^2+(3-0)^2=25, so (2-h)^2=16, giving 2-h=+-4, i.e. h=-2 or h=6.
✦ Working through each part gives: (x+2)^2+y^2=25, or (x-6)^2+y^2=25.
Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.
Hint. A circle through the origin with x-intercept a and y-intercept b also passes through (a,0) and (0,b). Use the general form and these three points.
Using x^2+y^2+2gx+2fy+c=0: passing through (0,0) gives c=0. Passing through (a,0): a^2+2ga=0, so g=-a/2. Passing through (0,b): b^2+2fb=0, so f=-b/2. Substituting back into the general form gives x^2+y^2-ax-by=0.
✦ Working through each part gives: x^2+y^2-ax-by=0.
Find the equation of a circle with centre (2,2) and passing through the point (4,5).
Hint. The radius is the distance from the centre to the given point on the circle.
r^2=(4-2)^2+(5-2)^2=4+9=13. Substituting into (x-h)^2+(y-k)^2=r^2 with h=2, k=2: (x-2)^2+(y-2)^2=13.
✦ Working through each part gives: (x-2)^2+(y-2)^2=13.
Does the point (-2.5, 3.5) lie inside, outside, or on the circle x^2+y^2=25?
Hint. Substitute the point's coordinates into the left-hand side x^2+y^2 and compare with 25.
(-2.5)^2+(3.5)^2=6.25+12.25=18.5. Since 18.5<25, the point is closer to the centre than the radius allows.
✦ Working through each part gives: The point lies inside the circle, since 18.5 < 25.
Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh110.pdf) — Exercise 10.1 (15 questions), Exercise 10.2 (12 questions), Exercise 10.3 (20 questions), Exercise 10.4 (15 questions), plus the chapter's Miscellaneous Exercise (8 questions), 70 questions total. Confirmed against the CBSE curriculum PDF that Conic Sections has no formative-only carve-out at all, unlike its neighbouring chapters Straight Lines and Introduction to Three-Dimensional Geometry — every phrase in its syllabus line is summative. Confirmed the book teaches the general circle equation only through completing the square (Example 3), never naming g, f, c as constants, and never mentions hyperbola asymptotes anywhere (zero hits for the word itself, and none of Exercise 10.4's 15 questions ask for one). Exercise 10.1 Q3's centre coordinates were cross-checked via a 300dpi page render after the raw PDF text extraction reversed their order. Several exercise answers (10.1 Q10/Q11, 10.4 Q15, Misc Q2/Q3/Q5/Q8) were cross-verified against known standard results for this exact problem set.. Questions are referenced from the NCERT textbook for identification.
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