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Circle question

2024 · 30 Jan · Shift 1 · Q33
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  5. /2024 · 30 Jan · Shift 1 · Q33

Circle question

2024 · 30 Jan · Shift 1 · Q33

JEE MainMathematicsCircleMCQ+4 / −1
If the circles (x+1)2+(y+2)2=r2(x+1)^2+(y+2)^2=r^2(x+1)2+(y+2)2=r2 and x2+y2−4x−4y+4=0x^2+y^2-4 x-4 y+4=0x2+y2−4x−4y+4=0 intersect at exactly two distinct points, then
  1. A
    12<r<7\frac{1}{2}\lt \mathrm{r}\lt 721​<r<7
  2. B
    3<r<73\lt \mathrm{r}\lt 73<r<7
  3. C
    5<r<95\lt \mathrm{r}\lt 95<r<9
  4. D
    0<r<70\lt \mathrm{r}\lt 70<r<7
View written solutionFree

Correct answer: B

  1. Identify centers and radii of the two circles

The first circle is

(x+1)2+(y+2)2=r2(x+1)^2+(y+2)^2=r^2(x+1)2+(y+2)2=r2

So its center is

C1=(−1,−2)C_1=(-1,-2)C1​=(−1,−2)

and radius is

r1=r.r_1=r.r1​=r.

The second circle is

x2+y2−4x−4y+4=0.x^2+y^2-4x-4y+4=0.x2+y2−4x−4y+4=0.

Complete squares:

(x2−4x)+(y2−4y)+4=0(x^2-4x)+(y^2-4y)+4=0(x2−4x)+(y2−4y)+4=0 (x−2)2−4+(y−2)2−4+4=0(x-2)^2-4+(y-2)^2-4+4=0(x−2)2−4+(y−2)2−4+4=0 (x−2)2+(y−2)2=4.(x-2)^2+(y-2)^2=4.(x−2)2+(y−2)2=4.

So its center is

C2=(2,2)C_2=(2,2)C2​=(2,2)

and radius is

r2=2.r_2=2.r2​=2.
  1. Find distance between centers
d=C1C2=(2−(−1))2+(2−(−2))2=32+42=5.d=C_1C_2=\sqrt{(2-(-1))^2+(2-(-2))^2} =\sqrt{3^2+4^2}=5.d=C1​C2​=(2−(−1))2+(2−(−2))2​=32+42​=5.
  1. Condition for two distinct points of intersection

Two circles intersect at exactly two distinct points when

∣r1−r2∣<d<r1+r2.|r_1-r_2|<d<r_1+r_2.∣r1​−r2​∣<d<r1​+r2​.

Substitute r1=rr_1=rr1​=r, r2=2r_2=2r2​=2, d=5d=5d=5:

∣r−2∣<5<r+2.|r-2|<5<r+2.∣r−2∣<5<r+2.

Now solve both inequalities.

  • From
5<r+25<r+25<r+2

we get

r>3.r>3.r>3.
  • From
∣r−2∣<5|r-2|<5∣r−2∣<5

we get

−5<r−2<5-5<r-2<5−5<r−2<5 −3<r<7.-3<r<7.−3<r<7.

Since radius must be positive, this is consistent, and combining with r>3r>3r>3 gives

3<r<7.3<r<7.3<r<7.
  1. Check options
  • A: 12<r<7\frac12<r<721​<r<7 → includes values like r=1r=1r=1, which do not satisfy intersection at two points. False.
  • B: 3<r<73<r<73<r<7 → correct.
  • C: 5<r<95<r<95<r<9 → misses valid values such as r=4r=4r=4. False.
  • D: 0<r<70<r<70<r<7 → includes values like r=1r=1r=1. False.

Therefore, the correct option is

B\boxed{B}B​

with

3<r<7.\boxed{3<r<7}.3<r<7​.
  1. Comparison with stored answer

Stored correct answer: B

My derived answer: B

They agree.

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