A circle with centre B on the x-axis passes through the points P(-2,-3) and R(5, -4). What is the equation of the circle?

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Answer 1 · 2018-04-24T17:58:30

# (x-2)^2+(y-0)^2=25, or, x^2+y^2-4x-21=0#.

The centre #B# of the reqd. circle, say #S#, lies on the X-Axis.

Any point on the X-Axis has y-co-ordinate zero.

So, let us have, #B=B(h,0)#.

Also, let #r# be the radius.

#P(-2,-3), R(5,-4) in S...[Given], &," centre "B(h,0)#.

# rArr BP^2=BR^2=r^2#.

#rArr (h+2)^2+(0-(-3))^2=(h-5)^2+(0-(-4))^2#.

#rArr (h^2+4h+4)+9=(h^2-10h+25)+16#.

# rArr 14h=28#.

# rArr h=2#.

Then, #r^2=(h+2)^2+9=(2+2)^2+9=25#.

Thus, we find that, for #S#, the centre #B(2,0), &," radius r=5"#.

Consequently, #S : (x-2)^2+(y-0)^2=25, or, #

#S : x^2+y^2-4x-21=0#.

Enjoy Maths.!