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Find the equation of a circle given by the points (-4,2),(-2,6) and (4,8)​

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Answer:

(x-3)²+(y-1)²=50

Explanation:

we know,

general eqn of circle passing through a point is

r²=(x-h)²+(y-k)²------($)

then,at (-4,2) the eqn becomes

r²=(-4-h)²+(2-k)²----(1)

at (-2,6),

r²=(-2-h)²+(6-k)²-----(2)

at (4,8),

r²=(4-h)²+(8-k)²-------(3)

Now,

from (1) and(2),

(-4-h)²+(2-k)²=(-2-h)²+(6-k)²

or,16+8h+h²+4-4k+k²=4+4h+h²+36-12k+k²

or,h²-h²+8h-4h+k²-k²-4k+12k+16+4-4-36=0

or,4h+8k-20=0

or,4(h+2k)=20

or,h+2k=5------(4)

also,from (2) and (3),

(-2-h)²+(6-k)²=(4-h)²+(8-k)²

or,4+4h+h²+36-12k+k²=16-8h+h²+64-16k+k²

or,h²-h²+4h+8h+k²-k²-12k+16k+4+36-64-16=0

or,12h+4k-40=0

or,4(3h+k)=40

or,3h+k=10-------(5)

Now,multiplying eqn (5)by 2 then subtracting from (4),we get

h+2k=5

6h+2k=20

- - -

_________

-5h=-15

.:h=3

putting value of h in (4),we get

3+2k=5

or,2k=2

.:k=1

Now,putting value of k and h in eqn(1),

r²=(-7)²+(1)²

or,r²=49+1

.:r²=50

Now putting value of h,k and r² in eqn($),we get

(x-3)²+(y-1)²=50,which is required eqn of circle

User Ismael Di Vita
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