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Circle question

2021 · 27 Jul · Shift 1 · Q38
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  5. /2021 · 27 Jul · Shift 1 · Q38

Circle question

2021 · 27 Jul · Shift 1 · Q38

JEE MainMathematicsCircleMCQ+4 / −1
Let A={(x,y)∈R×R∣2x2+2y2−2x−2y=1}A = \{ (x,y) \in R \times R|2{x^2} + 2{y^2} - 2x - 2y = 1\}A={(x,y)∈R×R∣2x2+2y2−2x−2y=1}, B={(x,y)∈R×R∣4x2+4y2−16y+7=0}B = \{ (x,y) \in R \times R|4{x^2} + 4{y^2} - 16y + 7 = 0\}B={(x,y)∈R×R∣4x2+4y2−16y+7=0} and C={(x,y)∈R×R∣x2+y2−4x−2y+5≤r2}C = \{ (x,y) \in R \times R|{x^2} + {y^2} - 4x - 2y + 5 \le {r^2}\}C={(x,y)∈R×R∣x2+y2−4x−2y+5≤r2}. Then the minimum value of |r| such that A∪B⊆CA \cup B \subseteq CA∪B⊆C is equal to
  1. A
    3+102{{3 + \sqrt {10} } \over 2}23+10​​
  2. B
    2+102{{2 + \sqrt {10} } \over 2}22+10​​
  3. C
    3+252{{3 + 2\sqrt 5 } \over 2}23+25​​
  4. D
    1+51 + \sqrt 51+5​
View written solutionFree

Correct answer: C

  1. Interpret the sets as circles/discs

We rewrite each equation in standard form.

Set AAA

Given 2x2+2y2−2x−2y=12x^2+2y^2-2x-2y=12x2+2y2−2x−2y=1 Divide by 222: x2+y2−x−y=12x^2+y^2-x-y=\frac12x2+y2−x−y=21​ Complete squares: x2−x+y2−y=12x^2-x+y^2-y=\frac12x2−x+y2−y=21​ (x−12)2−14+(y−12)2−14=12\left(x-\frac12\right)^2-\frac14+\left(y-\frac12\right)^2-\frac14=\frac12(x−21​)2−41​+(y−21​)2−41​=21​ (x−12)2+(y−12)2=1\left(x-\frac12\right)^2+\left(y-\frac12\right)^2=1(x−21​)2+(y−21​)2=1 So AAA is the circle with center OA=(12,12),RA=1O_A=\left(\frac12,\frac12\right), \quad R_A=1OA​=(21​,21​),RA​=1

Set BBB

Given 4x2+4y2−16y+7=04x^2+4y^2-16y+7=04x2+4y2−16y+7=0 Divide by 444: x2+y2−4y+74=0x^2+y^2-4y+\frac74=0x2+y2−4y+47​=0 Complete square in yyy: x2+(y−2)2−4+74=0x^2+(y-2)^2-4+\frac74=0x2+(y−2)2−4+47​=0 x2+(y−2)2=94x^2+(y-2)^2=\frac94x2+(y−2)2=49​ So BBB is the circle with center OB=(0,2),RB=32O_B=(0,2), \quad R_B=\frac32OB​=(0,2),RB​=23​

Set CCC

Given x2+y2−4x−2y+5≤r2x^2+y^2-4x-2y+5\le r^2x2+y2−4x−2y+5≤r2 Complete squares: x2−4x+y2−2y+5≤r2x^2-4x+y^2-2y+5\le r^2x2−4x+y2−2y+5≤r2 (x−2)2−4+(y−1)2−1+5≤r2 (x-2)^2-4+(y-1)^2-1+5\le r^2(x−2)2−4+(y−1)2−1+5≤r2 (x−2)2+(y−1)2≤r2 (x-2)^2+(y-1)^2\le r^2(x−2)2+(y−1)2≤r2 So CCC is the disc centered at OC=(2,1)O_C=(2,1)OC​=(2,1) with radius ∣r∣|r|∣r∣.

We need the minimum value of ∣r∣|r|∣r∣ such that both circles AAA and BBB lie entirely inside this disc.

  1. Condition for a circle to lie inside the disc CCC

If a circle has center at distance ddd from OCO_COC​ and radius RRR, then it is fully contained in CCC iff ∣r∣≥d+R|r|\ge d+R∣r∣≥d+R So we compute this for AAA and BBB.

  1. Containment of AAA inside CCC

Distance between centers:

=\sqrt{\left(\frac32\right)^2+\left(\frac12\right)^2} =\sqrt{\frac94+\frac14} =\sqrt{\frac{10}{4}} =\frac{\sqrt{10}}{2}$$ Hence required radius to contain $A$ is $$d_A+R_A=\frac{\sqrt{10}}{2}+1=\frac{2+\sqrt{10}}{2}$$ --- 4. **Containment of $B$ inside $C$** Distance between centers: $$d_B=\sqrt{(2-0)^2+(1-2)^2}=\sqrt{4+1}=\sqrt5$$ Hence required radius to contain $B$ is $$d_B+R_B=\sqrt5+\frac32=\frac{3+2\sqrt5}{2}$$ --- 5. **Containment of $A\cup B$ inside $C$** To contain both, we need $$|r|\ge \max\left(\frac{2+\sqrt{10}}{2},\frac{3+2\sqrt5}{2}\right)$$ Now compare: $$\frac{2+\sqrt{10}}{2}\approx \frac{2+3.162}{2}\approx 2.581$$ $$\frac{3+2\sqrt5}{2}\approx \frac{3+4.472}{2}\approx 3.736$$ So the minimum value is $$|r|_{\min}=\frac{3+2\sqrt5}{2}$$ --- 6. **Option check** This matches **Option C**.
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