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find the equation of a circle touching lines x=-1,x=3 and y=3

find the equation of a circle touching lines x=-1,x=3 and y=3

Grade:

3 Answers

Arpit Jaiswal
31 Points
8 years ago

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

Pranay Shah
18 Points
8 years ago

equations:

(x-2)^2+(y-1)^2=1

(x-4)^2+(y-1)^2=1

 

the question is solved by simple geometry.

ankitesh gupta
63 Points
8 years ago

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

 plzzzzzzzzz approve it if it helped you by clicking yes button

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