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

2024 · 27 Jan · Shift 2 · Q58
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. /2024 · 27 Jan · Shift 2 · Q58

Circle question

2024 · 27 Jan · Shift 2 · Q58

JEE MainMathematicsCircleNumerical+4 / −1
Consider a circle (x−α)2+(y−β)2=50(x-\alpha)^2+(y-\beta)^2=50(x−α)2+(y−β)2=50, where α,β>0\alpha, \beta\gt 0α,β>0. If the circle touches the line y+x=0y+x=0y+x=0 at the point PPP, whose distance from the origin is 424 \sqrt{2}42​, then (α+β)2(\alpha+\beta)^2(α+β)2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 100

  1. The circle is
(x−α)2+(y−β)2=50(x-\alpha)^2+(y-\beta)^2=50(x−α)2+(y−β)2=50

So its center is

C(α,β)C(\alpha,\beta)C(α,β)

and its radius is

r=50=52.r=\sqrt{50}=5\sqrt{2}.r=50​=52​.
  1. The circle touches the line
x+y=0.x+y=0.x+y=0.

If it touches this line at point PPP, then the radius to the point of contact is perpendicular to the line.

Also, PPP lies on the line x+y=0x+y=0x+y=0 and its distance from the origin is 424\sqrt{2}42​.

Let

P=(t,−t).P=(t,-t).P=(t,−t).

Then its distance from origin is

t2+(−t)2=2t2=∣t∣2.\sqrt{t^2+(-t)^2}=\sqrt{2t^2}=|t|\sqrt{2}.t2+(−t)2​=2t2​=∣t∣2​.

Given this equals 424\sqrt{2}42​,

∣t∣=4.|t|=4.∣t∣=4.

So possible points are

(4,−4)or(−4,4).(4,-4) \quad \text{or} \quad (-4,4).(4,−4)or(−4,4).
  1. Since α,β>0\alpha,\beta>0α,β>0, the center lies in the first quadrant. The radius at the point of contact must be along the normal to the line x+y=0x+y=0x+y=0, whose direction is (1,1)(1,1)(1,1).

From P=(−4,4)P=(-4,4)P=(−4,4), moving in direction (1,1)(1,1)(1,1) enters the first quadrant, while from (4,−4)(4,-4)(4,−4) it does not give both coordinates positive for the center at distance 525\sqrt{2}52​.

Hence,

P=(−4,4).P=(-4,4).P=(−4,4).
  1. A unit normal vector to the line x+y=0x+y=0x+y=0 is
12(1,1).\frac{1}{\sqrt{2}}(1,1).2​1​(1,1).

Since radius =52=5\sqrt{2}=52​, the vector from PPP to center is

52⋅12(1,1)=5(1,1)=(5,5).5\sqrt{2}\cdot \frac{1}{\sqrt{2}}(1,1)=5(1,1)=(5,5).52​⋅2​1​(1,1)=5(1,1)=(5,5).

Therefore,

C=P+(5,5)=(−4,4)+(5,5)=(1,9).C=P+(5,5)=(-4,4)+(5,5)=(1,9).C=P+(5,5)=(−4,4)+(5,5)=(1,9).

So,

α=1,β=9.\alpha=1,\quad \beta=9.α=1,β=9.
  1. Hence,
(α+β)2=(1+9)2=102=100.(\alpha+\beta)^2=(1+9)^2=10^2=100.(α+β)2=(1+9)2=102=100.

Therefore, the required integer is

100.\boxed{100}.100​.
  1. Comparison with stored answer: Stored correct answer = 100100100. Our derived answer is also 100100100, so they agree.
PreviousNext

More from Circle

  • Equations of two diameters of a circle are 2x−3y=5 and 3x−4y=7. The line joining the points (−722​,−4) and (−71​,3) intersects the circle at only one point P(α,β). Then, 17β−α…2024 · Numerical
  • If the circles (x+1)2+(y+2)2=r2 and x2+y2−4x−4y+4=0 intersect at exactly two distinct points, then2024 · MCQ
  • Consider two circles C1​:x2+y2=25 and C2​:(x−α)2+y2=16, where α∈(5,9). Let the angle between the two radii (one to each circle) drawn from one of the intersection points of C1​ and C2​ be sin−1(863​​)…2024 · Numerical
  • If one of the diameters of the circle x2+y2−10x+4y+13=0 is a chord of another circle C, whose center is the point of intersection of the lines 2x+3y=12 and 3x−2y=5, then the radius of the circle C is :2024 · MCQ
  • Let a variable line passing through the centre of the circle x2+y2−16x−4y=0, meet the positive co-ordinate axes at the points A and B. Then the minimum value of OA+OB, where O is the origin, is equal to2024 · MCQ
  • Let the point (p,p+1) lie inside the region E={(x,y):3−x≤y≤9−x2​,0≤x≤3}. If the set of all values of p is the interval (a,b), then b2+b−a2 is equal to ​…2023 · Numerical
  • A circle passing through the point P(α,β) in the first quadrant touches the two coordinate axes at the points A and B. The point P is above the line AB. The point Q on the line segment AB is the foot of…2023 · Numerical
  • Consider a circle C1​:x2+y2−4x−2y=α−5. Let its mirror image in the line y=2x+1 be another circle C2​:5x2+5y2−10fx−10gy+36=0. Let r be the radius of C2​. Then α+r is equal to ​…2023 · Numerical