Question

# Let A={(x,y)∈R×R|2x2+2y2−2x−2y=1}, B={(x,y)∈R×R|4x2+4y2−16y+7=0} and C={(x,y)∈R×R|x2+y2−4x−2y+5≤r2}. Then the minimum value of |r| such that A∪B⊆C is equal to:

A
2+102
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B
3+252
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C
1+5
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D
3+102
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Solution

## The correct option is B 3+2√52A={(x,y)∈R×R|2x2+2y2−2x−2y=1}, Equation of circle is x2+y2−x−y−12=0 Centre of circle, C1≡(12,12) and radius, r1=1 B={(x,y)∈R×R|4x2+4y2−16y+7=0} Equation of circle is x2+y2−4y+74=0 Centre of circle, C2≡(0,2) and radius, r2=32 C={(x,y)∈R×R|x2+y2−4x−2y+5≤r2} Equation of circle is (x−2)2+(y−1)2≤r2 Centre of circle, C3≡(2,1) and radius, r3=|r| From diagram it is clear that C3C2≤∣∣∣r−32∣∣∣ ⇒∣∣∣r−32∣∣∣≥√5 Case (1): r−32≥√5 ⇒r≥3+2√52 Case (2): r−32≤−√5 ⇒r≤3−2√52 Which is not possible. ∴rmin=3+2√52

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