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

2023 · 12 Apr · Shift 1 · Q37
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. /2023 · 12 Apr · Shift 1 · Q37

Circle question

2023 · 12 Apr · Shift 1 · Q37

JEE MainMathematicsCircleNumerical+4 / −1
Two circles in the first quadrant of radii r1r_{1}r1​ and r2r_{2}r2​ touch the coordinate axes. Each of them cuts off an intercept of 2 units with the line x+y=2x+y=2x+y=2. Then r12+r22−r1r2r_{1}^{2}+r_{2}^{2}-r_{1} r_{2}r12​+r22​−r1​r2​ is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 7

  1. Equation of the circles touching both coordinate axes

A circle in the first quadrant touching both coordinate axes has center at (r,r)(r,r)(r,r) and radius rrr.

So its equation is

(x−r)2+(y−r)2=r2.(x-r)^2+(y-r)^2=r^2.(x−r)2+(y−r)2=r2.

If the two circles have radii r1r_1r1​ and r2r_2r2​, their centers are (r1,r1)(r_1,r_1)(r1​,r1​) and (r2,r2)(r_2,r_2)(r2​,r2​).

  1. Condition involving the line x+y=2x+y=2x+y=2

We are told that each circle cuts off an intercept of length 222 units on the line x+y=2x+y=2x+y=2.

For a circle, if a line at perpendicular distance ddd from the center cuts a chord of length LLL, then

L=2r2−d2.L=2\sqrt{r^2-d^2}.L=2r2−d2​.

Here, L=2L=2L=2, so

2r2−d2=2  ⟹  r2−d2=1  ⟹  r2−d2=1.2\sqrt{r^2-d^2}=2 \implies \sqrt{r^2-d^2}=1 \implies r^2-d^2=1.2r2−d2​=2⟹r2−d2​=1⟹r2−d2=1.
  1. Distance from center to the line

The line is

x+y−2=0.x+y-2=0.x+y−2=0.

Distance of center (r,r)(r,r)(r,r) from this line is

d=∣r+r−2∣12+12=∣2r−2∣2=2 ∣r−1∣.d=\frac{|r+r-2|}{\sqrt{1^2+1^2}}=\frac{|2r-2|}{\sqrt{2}}=\sqrt{2}\,|r-1|.d=12+12​∣r+r−2∣​=2​∣2r−2∣​=2​∣r−1∣.

Hence

d2=2(r−1)2.d^2=2(r-1)^2.d2=2(r−1)2.

Using the chord condition,

r2−2(r−1)2=1.r^2-2(r-1)^2=1.r2−2(r−1)2=1.

Now simplify:

r2−2(r2−2r+2)=1r^2-2(r^2-2r+2)=1r2−2(r2−2r+2)=1 r2−2r2+4r−4=1r^2-2r^2+4r-4=1r2−2r2+4r−4=1 −r2+4r−5=1-r^2+4r-5=1−r2+4r−5=1 −r2+4r−6=0-r^2+4r-6=0−r2+4r−6=0 r2−4r+6=0.r^2-4r+6=0.r2−4r+6=0.

This gives discriminant

Δ=16−24=−8<0,\Delta=16-24=-8<0,Δ=16−24=−8<0,

which is impossible. So let us carefully interpret the statement.

  1. Correct interpretation of “cuts off an intercept of 2 units with the line”

The line segment intercepted by the circle on the line x+y=2x+y=2x+y=2 is a chord of length 222. If we parameterize the line:

y=2−x.y=2-x.y=2−x.

Substitute in the circle equation:

(x−r)2+(2−x−r)2=r2.(x-r)^2+(2-x-r)^2=r^2.(x−r)2+(2−x−r)2=r2.

Expand:

(x−r)2+(2−r−x)2=r2.(x-r)^2+(2-r-x)^2=r^2.(x−r)2+(2−r−x)2=r2.

This is a quadratic in xxx, whose two roots give the intersection points. Since along the line x+y=2x+y=2x+y=2, distance between two points corresponds to 2\sqrt{2}2​ times the difference in their xxx-coordinates, if the chord length is 222, then difference of roots is

22=2.\frac{2}{\sqrt{2}}=\sqrt{2}.2​2​=2​.

Now compute the quadratic.

First,

(x−r)2=x2−2rx+r2,(x-r)^2=x^2-2rx+r^2,(x−r)2=x2−2rx+r2, (2−r−x)2=x2−2(2−r)x+(2−r)2.(2-r-x)^2=x^2-2(2-r)x+(2-r)^2.(2−r−x)2=x2−2(2−r)x+(2−r)2.

So

2x2−2rx−2(2−r)x+r2+(2−r)2=r2.2x^2-2r x-2(2-r)x+r^2+(2-r)^2=r^2.2x2−2rx−2(2−r)x+r2+(2−r)2=r2. 2x2−2[r+(2−r)]x+(2−r)2=02x^2-2[r+(2-r)]x+(2-r)^2=02x2−2[r+(2−r)]x+(2−r)2=0 2x2−4x+(2−r)2=0.2x^2-4x+(2-r)^2=0.2x2−4x+(2−r)2=0.

For quadratic ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0, difference of roots is

Da,D=b2−4ac.\frac{\sqrt{D}}{a}, \quad D=b^2-4ac.aD​​,D=b2−4ac.

Here,

a=2,b=−4,c=(2−r)2.a=2,\quad b=-4,\quad c=(2-r)^2.a=2,b=−4,c=(2−r)2.

So

D=16−8(2−r)2.D=16-8(2-r)^2.D=16−8(2−r)2.

Difference of roots is

16−8(2−r)22.\frac{\sqrt{16-8(2-r)^2}}{2}.216−8(2−r)2​​.

This must equal 2\sqrt{2}2​:

16−8(2−r)22=2.\frac{\sqrt{16-8(2-r)^2}}{2}=\sqrt{2}.216−8(2−r)2​​=2​.

Squaring,

16−8(2−r)2=816-8(2-r)^2=816−8(2−r)2=8 8(2−r)2=88(2-r)^2=88(2−r)2=8 (2−r)2=1.(2-r)^2=1.(2−r)2=1.

Thus

r=1orr=3.r=1 \quad \text{or} \quad r=3.r=1orr=3.

So the two radii are

r1=1,r2=3.r_1=1,\quad r_2=3.r1​=1,r2​=3.
  1. Required value

We need

r12+r22−r1r2=12+32−(1)(3)=1+9−3=7.r_1^2+r_2^2-r_1r_2=1^2+3^2-(1)(3)=1+9-3=7.r12​+r22​−r1​r2​=12+32−(1)(3)=1+9−3=7.
  1. Comparison with stored answer

Derived answer is

7,7,7,

which matches the stored correct answer.

PreviousNext

More from Circle

  • The locus of the mid points of the chords of the circle C1​:(x−4)2+(y−5)2=4 which subtend an angle θi​ at the centre of the circle C1​, is a circle of radius ri​. If θ1​=3π​,θ3​=32π​…2023 · MCQ
  • The points of intersection of the line ax+by=0,(aeb) and the circle x2+y2−2x=0 are A(α,0) and B(1,β). The image of the circle with AB as a diameter in the line x+y+2=0 is :2023 · MCQ
  • Let the tangents at the points A(4,−11) and B(8,−5) on the circle x2+y2−3x+10y−15=0, intersect at the point C. Then the radius of the circle, whose centre is C and the line joining A and B is its tangent, is equal to :2023 · MCQ
  • Let P(a1​,b1​) and Q(a2​,b2​) be two distinct points on a circle with center C(2​,3​). Let O be the origin and OC be perpendicular to both CP and CQ. If…2023 · Numerical
  • The set of all values of a2 for which the line x+y=0 bisects two distinct chords drawn from a point P(21+a​,21−a​) on the circle 2x2+2y2−(1+a)x−(1−a)y=0, is equal to :2023 · MCQ
  • Let a circle C : (x − h)2 + (y − k)2 = r2, k > 0, touch the x-axis at (1, 0). If the line x + y = 0 intersects the circle C at P and Q such that the length of the chord PQ is 2, then the value of h + k + r is equal to ​…2022 · Numerical
  • Let the abscissae of the two points P and Q be the roots of 2x2−rx+p=0 and the ordinates of P and Q be the roots of x2−sx−q=0. If the equation of the circle described on PQ as diameter is 2(x2+y2)−11x−14y−22=0…2022 · Numerical
  • Let the abscissae of the two points P and Q on a circle be the roots of x2−4x−6=0 and the ordinates of P and Q be the roots of y2+2y−7=0. If PQ is a diameter of the circle x2+y2+2ax+2by+c=0…2022 · MCQ