To prove that
L1 : 2x+5y+11=0
is a tangent to the circle
C : x²+y²+2x-8y-12=0
it is sufficient to prove that the distance of the centre O of the circle C to the line L1 equals the radius of the circle.
To find the centre and radius of the circle C, we complete squares,
C : x²+y²+2x-8y-12=0
or
C : (x+1)²+(y-4)²=12+1+16=29
or
centre: O(-1,4)
radius: √29.
Distance of O from L1:
D=(2(-1)+5(4)+11)/√(2^2+5^2)
=29/√29
=√29
Since distance of centre O from line L1 equals the radius of circle C, L1 is tangent to C.
Show that the line 2x+5y+11=0 is a Tangent to the circle with equation x²+y²+2x-8y-12=0. Then find the tangent points.
Show step pl
2 answers
you want to find where
2x+5y+11=0
intersects
x²+y²+2x-8y-12=0
just doing a straight substitution, we get
4x^2+4y^2+8x-32y-48 = 0
2x = -(5y+11)
so,
(2x)^2+4y^2+4(2x)-32y-48 = 0
(5y+11)^2 + 4y^2 - 4(5y+11) - 32y - 48 = 0
29y^2+58y+29 = 0
29(y+1)^2 = 0
y = -1
Since there is only one solution, the line intersects the circle in only one point: it is tangent to the circle.
2x+5y+11=0
intersects
x²+y²+2x-8y-12=0
just doing a straight substitution, we get
4x^2+4y^2+8x-32y-48 = 0
2x = -(5y+11)
so,
(2x)^2+4y^2+4(2x)-32y-48 = 0
(5y+11)^2 + 4y^2 - 4(5y+11) - 32y - 48 = 0
29y^2+58y+29 = 0
29(y+1)^2 = 0
y = -1
Since there is only one solution, the line intersects the circle in only one point: it is tangent to the circle.
2 answers