amaths

1.Find the equation of the circle with the end-points of its diameter at A(-4,3) and B(8,7)

centre = mid-point = ( (-4 + 8)/2, (3 + 7)/2 ) = (2,5)
radius
= sqrt( (8-2)^2 + (7-5)^2 ) = sqrt(40)
equation of the circle
(x - 2)^2 + (y - 5)^2 = 40
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2.Find the equation of the circle with its centre at(2,4) which touches the line x+y-3=0
radius
= (2+4-3)/sqrt(1 + 1)
= 3/sqrt(2)
equation of the circle
(x - 2)^2 + (y - 4)^2 = 9/2

1.(x+4)(x-8)+(y-3)(y-7)=0
x^2+y^2-4x-10y-11=0

2.because circle touches the line
∴r =circle and the line displacement
=∣(2+4-3)/√(1+1)∣
=3/√2
(x-2)^2+(y-4)^2=(3/√2)^2
x^2+y^2-4x-8y+4+16=9/2
x^2+y^2-4x-8y+15.5=0
2x^2+2y^2-8x-16y+31=0