Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Circle question

2025 · 3 Apr · Shift 2 · Q45
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Circle
  5. /2025 · 3 Apr · Shift 2 · Q45

Circle question

2025 · 3 Apr · Shift 2 · Q45

JEE MainMathematicsCircleMCQ+4 / −1
If the four distinct points (4,6),(−1,5),(0,0)(4,6),(-1,5),(0,0)(4,6),(−1,5),(0,0) and (k,3k)(k, 3 k)(k,3k) lie on a circle of radius rrr, then 10k+r210 k+r^210k+r2 is equal to
  1. A
    34
  2. B
    32
  3. C
    35
  4. D
    33
View written solutionFree

Correct answer: C

  1. Write the general equation of the circle

A circle can be written as x2+y2+Dx+Ey+F=0.x^2+y^2+Dx+Ey+F=0.x2+y2+Dx+Ey+F=0. We use the three given points (4,6),(−1,5),(0,0)(4,6),(-1,5),(0,0)(4,6),(−1,5),(0,0) to determine D,E,FD,E,FD,E,F.

  1. Substitute the point (0,0)(0,0)(0,0)

02+02+D(0)+E(0)+F=0  ⟹  F=0.0^2+0^2+D(0)+E(0)+F=0 \implies F=0.02+02+D(0)+E(0)+F=0⟹F=0.

So the circle is x2+y2+Dx+Ey=0.x^2+y^2+Dx+Ey=0.x2+y2+Dx+Ey=0.

  1. Substitute the point (4,6)(4,6)(4,6)

42+62+4D+6E=04^2+6^2+4D+6E=042+62+4D+6E=0 16+36+4D+6E=016+36+4D+6E=016+36+4D+6E=0 52+4D+6E=052+4D+6E=052+4D+6E=0 2D+3E=−26.(1)2D+3E=-26. \quad (1)2D+3E=−26.(1)

  1. Substitute the point (−1,5)(-1,5)(−1,5)

(−1)2+52−D+5E=0(-1)^2+5^2-D+5E=0(−1)2+52−D+5E=0 1+25−D+5E=01+25-D+5E=01+25−D+5E=0 26−D+5E=026-D+5E=026−D+5E=0 −D+5E=−26.(2)-D+5E=-26. \quad (2)−D+5E=−26.(2)

  1. Solve for DDD and EEE

From (2): D=26+5E.D=26+5E.D=26+5E.

Substitute into (1): 2(26+5E)+3E=−262(26+5E)+3E=-262(26+5E)+3E=−26 52+10E+3E=−2652+10E+3E=-2652+10E+3E=−26 13E=−7813E=-7813E=−78 E=−6.E=-6.E=−6.

Then D=26+5(−6)=26−30=−4.D=26+5(-6)=26-30=-4.D=26+5(−6)=26−30=−4.

Hence the circle is x2+y2−4x−6y=0.x^2+y^2-4x-6y=0.x2+y2−4x−6y=0.

  1. Find the fourth point condition

The point (k,3k)(k,3k)(k,3k) lies on the circle, so substitute:

k2+(3k)2−4k−6(3k)=0k^2+(3k)^2-4k-6(3k)=0k2+(3k)2−4k−6(3k)=0 k2+9k2−4k−18k=0k^2+9k^2-4k-18k=0k2+9k2−4k−18k=0 10k2−22k=010k^2-22k=010k2−22k=0 2k(5k−11)=0.2k(5k-11)=0.2k(5k−11)=0.

Thus k=0ork=115.k=0 \quad \text{or} \quad k=\frac{11}{5}.k=0ork=511​.

But (0,0)(0,0)(0,0) is already one of the given points, and the four points are distinct, so k≠0.k \ne 0.k=0. Therefore, k=115.k=\frac{11}{5}.k=511​.

  1. Find the radius rrr

For the circle x2+y2−4x−6y=0,x^2+y^2-4x-6y=0,x2+y2−4x−6y=0, complete the square:

x2−4x+y2−6y=0x^2-4x+y^2-6y=0x2−4x+y2−6y=0 (x−2)2−4+(y−3)2−9=0(x-2)^2-4+(y-3)^2-9=0(x−2)2−4+(y−3)2−9=0 (x−2)2+(y−3)2=13.(x-2)^2+(y-3)^2=13.(x−2)2+(y−3)2=13.

So r2=13.r^2=13.r2=13.

  1. Compute 10k+r210k+r^210k+r2

10k+r2=10(115)+13=22+13=35.10k+r^2=10\left(\frac{11}{5}\right)+13=22+13=35.10k+r2=10(511​)+13=22+13=35.

  1. Compare with the options

The value is 35.\boxed{35}.35​. So the correct option is C.

  1. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.

PreviousNext

More from Circle

  • Let C be the circle x2+(y−1)2=2,E1​ and E2​ be two ellipses whose centres lie at the origin and major axes lie on x -axis and y -axis respectively. Let the straight line x+y=3 touch the curves C,E1​ and E2​ at P(x1​,y1​),Q(x2​,y2​)…2025 · Numerical
  • Let C1​ be the circle in the third quadrant of radius 3 , that touches both coordinate axes. Let C2​ be the circle with centre (1,3) that touches C1​ externally at the point (α,β). If (β−α)2=nm​…2025 · MCQ
  • A circle C of radius 2 lies in the second quadrant and touches both the coordinate axes. Let r be the radius of a circle that has centre at the point (2,5) and intersects the circle C at exactly two points. If the set of all possible…2025 · MCQ
  • Let the circle C touch the line x−y+1=0, have the centre on the positive x-axis, and cut off a chord of length 13​4​ along the line −3x+2y=1. Let H be the hyperbola α2x2​−β2y2​=1,…2025 · Numerical
  • Let circle C be the image of x2+y2−2x+4y−4=0 in the line 2x−3y+5=0 and A be the point on C such that OA is parallel to x-axis and A lies on the right hand side of the centre O of C. If B(α,β), with β<4…2025 · MCQ
  • Let the equation of the circle, which touches x-axis at the point (a,0),a>0 and cuts off an intercept of length b on y−axis be x2+y2−αx+βy+γ=0. If the circle lies below x−axis, then the ordered pair…2025 · MCQ
  • Let the line x+y=1 meet the circle x2+y2=4 at the points A and B. If the line perpendicular to AB and passing through the mid-point of the chord AB intersects the circle at C and D, then the area of the quadrilateral ABCD is equal to :2025 · MCQ
  • Let a circle C pass through the points (4, 2) and (0, 2), and its centre lie on 3x + 2y + 2 = 0. Then the length of the chord, of the circle C, whose mid-point is (1, 2), is:2025 · MCQ