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Question

The circles x2+y2−4x+4y+4=0 and x2+y2−4x−4y=0

A
Do not intersect
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B
Are not orthogonal
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C
Intersect orthogonally
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D
Concentric
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Solution

The correct option is B Are not orthogonal
S1=x2+y2+4y4x+4=0
S2=x2+y24x4y=0
S1S2=(x2+y24x+4y+4)(x2+y24x4y)
=x2+y24x+4y4x2y2+4x+4y
=8y4
8y4=0y=1/2
from S1=x2+y2+4y4x+4=0
C1 centre (2,-2)
R1 Radius = 4+44=2
similarly S2=X2+y24x4y=0
C2 centre (2,2)
R2 Radius =4+4=8=22
Distance between centre of circles,
|C1C2|=(22)2+(2+2)2=0+42=42=4
Now R1+R2=2+22=2(1+2)|C1C2|
R1+R2>
The condition for orthogonality =2gg+2ff=c+c
2(2)(2)+2(2)(2)=4+0
88=04
So, no circles do not intersect orthogonality,
Option (b) is correct

1169582_1139999_ans_0484c9d22e3341769d5a59d02a0f6123.jpeg

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