Question

if the circle x2+y2-8x+4y+c=0 touches the x axis then c=?​

Answer 1

Answer:

(4,-2)

Step-by-step explanation:

general form of equation of circle is

x²+y²+2gx+2fy+c=0

x²+y²-8x+4y+c=0

compare both equations

2g= -8

g= -8/2

g = -4

2f = 4

f = 4/2

f = 2

centre of circle is

(-g,-f) then

(-(-2),-2)

(4,-2)

Answer 2

Answer:

c=16

Step-by-step explanation:

x²+y²-8x+4y+c=0

standard eq. of circle: x²+y²+2gx+2fy+c=o

center of circle: C(-g,-f) =C(4,-2)

∵circle touches x-axis only one case is possible with center(4,-2) where circumference touches x-axis on (4,0) so radius will be 2 as center is(4,-2) [make a diagram to understand better]

method1

(radius)R=$\sqrt[]{ f^2+g^2-c}$

2=$\sqrt[]{16+4-c}$

on squaring both side

4=20-c

c=16

method2

(x-4)²+(y-2)²=2² is eq. of given circle

on solving we get

x²+y²-8x+4y+16=0

on comparing above eq. with given eq. c=16