Question

find the centre of the circle passing through the point (5,7) (6,6) (2.-2) and its radius also find​

Answer 1

let P(h,k) be the centre of the circle and A(5,7) and B(6,6) and C(2,-2) be the points of the circle

then PA=PB=PC

from PA=PB we get

√(h-5)² + (k-7)² = √(h-6)²+(k-6)²

squaring both sides

(h-5)² + (k-7)² = (h-6)²+(k-6)²

h²-10h+25+k²-14k+49 = h²+36-12h+k²+36-12h

2h-2k=-2

dividing while eq by 2 we get

h-k= -1...(1)

now PA=PC give

√(h-5)²+(k-7)²= √(h-2)²+(k+2)²

s.b.s both sides

(h-5)²+(k-7)²= (h-2)²+(k+2)²

h²-10h+25+k²-14k+49 = h²-4h+4+k²+4k+4

on solving we get

-6h-18k = -66

dividing whole eq by 3 we get

-h-3k=-11

taking minus sign (-)common in LHL side

-(h+3k) =-11

h+3k=11..(2)

subtracting eq 1 from 2 we get,

h - k = -1

h+3k= 11

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0-4k= -12

4k=12

k=3

putting valu of k in eq 1

h-(3)=. -1

h = 2

thus the centre P(2,3)