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find the equation of a circle touching lines x=-1,x=3 and y=3
there is a length method, and there is a shorcut-
1)this is longer one, i will tell you the approach but not solve it- here we see these lines as tangents, and use equation of tangents of circles, and distance formulae.
2)this is short method, for this type of question with parallel lines given-
let the center be h,k
distance between x=-1 and x=3 is 4 units, as they are tangents, on opposite ends of diameter, therefore 4 units is length of diameter, and radius is 2 units.
also line through between these line, is the line on which center of the circle lies- i.e. x=1,
therfore x - coordinate of center is 1
i.e. h=1,
also , perpendicular distance of h,k from y=3 is 2, i.e h,k can lie on any of the two following lines-
y=5 or y=1,
i.e k=1 or k=5
therefore centre is (1,1) or (1,5) and radius is 2
therefore two possible equation of circles are-
(x-1)^2 + (y-1)^2=4 and, (x-1)^2+(y-5)^2=4
one on right side of the line, y=3 and other on left side.
hope this helps.
AJ
equations:
(x-2)^2+(y-1)^2=1
(x-4)^2+(y-1)^2=1
the question is solved by simple geometry.
CONSIDER A CENTER (-g,-f) NOW DISTANCE OF X=-1 FROM CENTER IS EQUAL TO DISTANCE OF X=3 FROM CENTER EQUAL TO DISTANCE OF Y=E FROM CENTER.
SO DISTANCE OF X=-1 FROM CENTER IS = MODULUS OF (-g*1-f*0+1)/1=DISTANCE OF X=3 FROM CENTER EQUAL=MODULUS OF (-g*1-f*0-3)/1
MODULUS OF(1-g)=MODULUS OF(-g-3) THEREFORE 1-g=(+ or -)(-g-3)
1-g = -g-3 or 1-g =g+3
1=-3(not possible) 2g=-2 g=-1.........(1) simillarly do for all you will get 2 value of f and 1 value of g them find R=moduluso= of (1-g) therefore R=2
now the equation in (x-g)2+(y-f)2=R2 you will get two equations and get your answer
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