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Coordinate geometry - Circles

For COMPETITION
Number of Total Problems: 2.
FOR PRINT ::: (Book)

Problem Num : 1

From : AMC10B

Type:

Section:Coordinate geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
Two circles of radius $2$ are centered at $(2,0)$ and at $(0,2).$ What is the area of the intersection of the interiors of the two circles?
$extbf{(A) } pi -2 qquad extbf{(B) } frac{pi}{2} qquad extbf{(C) } frac{pi sqrt{3}}{3} qquad extbf{(D) } 2(pi -...$
'

Category Circles

Analysis

Solution/Answer
You can find the area of half the intersection by subtracting the isosceles triangle in the sector from the whole sector. This sector is a quarter of the circle with radius $2,$ and the isosceles triangle is a right triangle. Therefore, the area of half the intersection is $frac{1}{4} 4pi - frac{1}{2}(2)(2) = pi - 2$ That means the area of the whole intersection is $oxed{mathrm{(D) } 2(pi-2)}$

Answer:

Problem Num : 2

From : AMC10

Type:

Section:Coordinate geometry

Theme:
Adjustment# : 0

Difficulty: 1

'
Which of the following describes the graph of the equation $(x+y)^2=x^2+y^2$ ?
$mathrm{(A) } extrm{the,empty,set}qquad mathrm{(B) } extrm{one,point}qquad mathrm{(C) } extrm{two,lines} ...$
'

Category Circles

Analysis

Solution/Answer
Expanding the left side, we have
$x^2+2xy+y^2=x^2+y^2Longrightarrow 2xy=0Longrightarrow xy=0Longrightarrow x = 0 extrm{ or } y = 0$
Thus there are two lines described in this graph, the horizontal line $y = 0$ and the vertical line $x=0$ . Thus, our answer is $mathrm{(C) }$ .

Answer:

Array ( [0] => 8076 [1] => 7830 ) 2