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Find the equation of the circle which passes through the point (-2,4) and through the points in which the circle x 2 +y 2 -2x -6y +6=0 is cut by the line 3x+2y-5=0

8 years ago

Answers : (1)

Aman Bansal
592 Points
							

Dear Devashish,

Hi, 
Generally speaking the center of the circle which equation is 
(x-a)^2+(y-b)^2=R^2 is the point M(a,b) and the radius of this circle is R 
(x – 2)^2 + (y + 3)^2 = 17 the center of this circle is A(2,-3)and the center of the circle (x – 5)^2 + y^2 = 32 is B(5,0)
The line (D): y=ax+b which passes through A and B is such that
A belongs to (D) ==> -3=a(2)+b=2a+b ==> b=-2a-3(*)
B belongs to (D) ==> 0=a(5)+b=5a+b ==>b=-5a(**)
(*) and (**) lead to -5a=-2a-3 ==> 5a-2a-3=0 ==> 3a-3=0 ==>3a=3
==> a=3/3=1 ==> a=1 From (**) b=-5a=-5(1)=-5
Finally the equation of (D) is y=x-5

The circle (C) with the equation (x-a)^2+(y-b)^2=r^2 has the same center with the circle which equation is (x – 4)^2 +(y + 3)^2 = 49 means a=4 and b=-3 O (4,-3) is the center of the circle (C). Therefore the equation of (C) (x – 4)^2 +(y + 3)^2=r^2 where r is the radius of (C) If (C) is tangent to the y-axis this (C) intersects the y-axis( which equation is x=0) at only one point M(x'', y'') x''=0 because M belongs to the y-axis M belongs to (C) y''=-3 because M is the orthogonal projection of O on the y-axis. The radius of (C) is the distance between O and M r=OM=sqrt[(x''-4)^2+(y''+3)^2]=sqrt[(0-4)^…
=sqrt[(-4)^2+(0)^2]=sqrt16=4 Finally the radius of (C) is r=4
Finally the equation of (C) is (x-4)^2+(y+3)^2=4^2=16

The area of the circle (C) is A =pi*r^2 where pi=3.14 and r is the radius of (C) A=3,14(4)^2=50.24

The area A'' of the circle (x-4)^2+(y+3)^2=49=7^2 is given by A''=pi*r''^2 where r'' is its radius . From the equation above we draw r=7 then A''=3.14(7)^2=153.86. The area S of the ring formed by the two circles is S= A''-A=7^2*pi-4^2*pi= 49pi-16pi=33pi units squared 

The line x=9 is secant to the circle (x – 6)^2 + (y + 2)^2 = 9 if by replacing x by its value in the equation of the circle we find two values of y i.e the equation (9-6)^2+(y+2)^2=9 has two roots then we will deduct that the line and the circle are secant (9-6)^2+(y+2)^2=9 ==> (3)^2 +(y+2)^2=9 
==> 9+(y+2)^2=9 => (y+2)^2=9-9=0 ==> (y+2)^2=0 ==> y+2=0
y=-2 One value for y therefore we can deduct that the line x=9 is tangent to the circle (x – 6)^2 + (y + 2)^2 = 9. If we had not found any value for y we would have deducted that the line and the circle are neither secant nor tangent

The circle (C) (x-a)^2+(y-b)^2=r^2 contain the points J(-6,0) K(-3,3) and L(0,0) 
J belongs to (C) ==> (-6-a)^2+(0-b)^2=r^2 ==> (-6-a)^2+(-b)^2=r^2 (*)
K belongs to (C) ==> (-3-a)^2+(3-b)^2=r^2 (**)
L belongs to (C) ==> (0-a)^2+(0-b)^2=r^2 ==> (-a)^2+(-b)^2=r^2 ==>
a^2+b^2=r^2 (***)

(*) ==> (-6-a)^2+(-b)^2=r^2 ==> 36-12a+a^2+b^2=r^2 ==> 
36-12a=-a^2-b^2+r^2=0 because from (***) a^2+b^2=r^2
Hence 36-12a=0 ==> 12a=36 ==> a=36/12=3 ==> a=3

(**) ==> (-3-a)^2+(3-b)^2=r^2 ==> 9-6a+a^2+9-6b+b^2=r^2 
==> 18-6a-6b+a^2+b^2=r^2 ==> 18-6a-6b=-a^2-b^2+r^2=0 
==> 18-6a-6b=0 By replacing a by its value we obtain 18-6(3)-6b=0
==. 18-18-6b=0 ==> -6b=0 ==> b=0
a^2+b^2=r^2 ==> 3^2+0^2=r^2 ==> 9=r^2 ==> r=-3 or r=3. Since r>=0 we deduct r=3
Finally the equation of the circle (C) is (x-3)^2+(y-0)^2=3^2
==> (x-3)^2+y^2=3^2

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Thanks

Aman Bansal

Askiitian Expert


8 years ago
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