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.
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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.
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.
To show that a line is a tangent to a circle, we need to find the point(s) of tangency.
Step 1: Find the center and radius of the circle.
Given the equation of the circle x²+y²+2x-8y-12=0, we can rewrite it in the standard form of a circle equation: (x - h)² + (y - k)² = r².
Comparing the given equation with the standard form, we can determine the center of the circle and the radius.
(x²+2x) + (y²-8y) = 12
Completing the square for x and y terms:
(x²+2x+1) + (y²-8y+16) = 12 + 1 + 16
(x+1)² + (y-4)² = 29
From this equation, we can determine that the center of the circle is at (-1, 4) and the radius is √29.
Step 2: Find the slope of the line.
The given line equation is 2x+5y+11=0. To find the slope of the line, we can rewrite the equation in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.
5y = -2x - 11
Divide by 5:
y = (-2/5)x - (11/5)
The slope of the given line is -2/5.
Step 3: Find the slope of the line perpendicular to the given line.
Since a tangent line is perpendicular to the radius of a circle at the point(s) of tangency, we need to find the slope of the line perpendicular to the given line: 2/5.
The slope of a line perpendicular to 2/5 can be found by taking the negative reciprocal: -1/(2/5) = -5/2.
Step 4: Find the equation of the line perpendicular to the given line through the center of the circle.
Using the point-slope form of a line (y - y1) = m(x - x1), where (x1, y1) is a point on the line and m is the slope:
(y - 4) = (-5/2)(x + 1)
Step 5: Find the intersection points of the line and the circle.
To find the points of tangency, we solve the system of equations formed by the line and the circle.
Substitute the equation of the perpendicular line into the equation of the circle:
(x+1)² + [(-5/2)(x + 1) + 4]² = 29
Simplify and solve the equation to find the values of x. Then substitute these values of x back into the equation of the perpendicular line to find the corresponding y-values.
Solving the equation will give us two values (x1, y1) and (x2, y2), which represent the points of tangency.
These steps show how to prove that the line 2x+5y+11=0 is a tangent to the circle x²+y²+2x-8y-12=0 and how to find the tangent points.