the problem solve itThe tangent line to a circle may be defined as the line that intersects the circle in a single point,called the point of tangency.If the equation if the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b,

the problem solve itThe tangent line to a circle may be defined as the line that intersects the circle in a single point,called the point of tangency.If the equation if the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b,
the problem solve it
The tangent line to a circle may be defined as the line that intersects the circle in a single point,called the point of tangency.If the equation if the circle is
x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b,
a) show that r^2(1 + m^2) = b^2
b) the point of tangency is (-r^2m/b,r^2/b)
c) the tangent line is perpendicular to the line containing the center of the circle and the point of tangency.
HINT:The quadratic equation x^2 + (mx + b) ^2 = r^2 has exactly one solution.

the problem solve itThe tangent line to a circle may be defined as the line that intersects the circle in a single point,called the point of tangency.If the equation if the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b,
Hi,
I assume you are using English system,so I will answer in English.
我假设你使用的是英文系统,所以我会用英语回答.
** according to the hint,X has exactly 1 solution,therefore
for the quadratic equation:x^2 + (mx + b) ^2 = r^2
x = (- B +/- (B^2 - 4AC) ^ (1/2) ) / 2A,
there is only one unique value of X,which implies:+/- (B^2-4AC) ^ (1/2) = 0
**equation 1:B^2-4AC = 0
**equation 2:x = ( -B +/- 0 ) / 2A -> x = -B / 2A
a)
find the formula for r in terms of m,x,and b:
y = mx + b
find the point of intersection,solve for x:
y^2 = (mx+b)^2 = m^2X^2+2mbx+b^2
use result in the equation for circle:
x^2 + y^2 = r^2
x^2 + ( m^2 x^2 + 2mb x + b^2) = r^2
(1+m^2) * X^2 + 2mb x + b^2 = r^2
(1+m^2)* X^2 + 2mb x + b^2 - r^2 = 0
use quadratic equation:
A = 1 + m^2
B = 2mb
C = b^2 - r^2
and use equation 1
**equation 1:B^2 - 4AC = 0
(2mb) ^2 - 4 (1 + m^2) * (b^2 - r^2) = 0
4m^2 b^2 - 4 (1+ m^2) * (b^2 - r^2) = 0
4m^2 b^2 - 4(b^2 - r^2 + b^2 m^2 -m^2 r^2) = 0
4m^2 b^2 - 4b^2 + 4 r^2 - 4m^2 b^2 + 4 m^2 r^2 = 0
- 4b^2 + 4 r^2 + 4 m^2 r^2 = 0
r^2 - b^2 + m^2r^2 = 0
r^2 + m^2r^2 = b^2
r^2 (1 + m^2) = b^2 *****
part a done" it is shown that r^2(1 + m^2) = b^2
b)
From part a,
r^2 (1 + m^2) = b^2
r^2 (1 + m^2) / b^2 = 1
r^2 / b^2 = 1 / (1+m^2)
**equation 3:1 / (1 + m^2) = r^2 / b^2
From part A,r^2(1+m^2) = b^2,
r^2 + r^2m^2 = b^2
r^2 = b^2 - r^2m^2
**equation 4:r^2 = b^2 - r^2m^2
now use **equation 2:x = -B / 2A
solve for x,given that there is only one unique x:
x = (- B)/ (2A) = -2mb / [ 2 * (1+m^2) ]
x = - mb * (1 / (1+m^2))
use **equation 3:1 / (1 + m^2) = r^2 / b^2
x = -mb * (r^2 / b^2)
x = -mr^2/b
x = - r^2 m / b
y = mx + b
= m * (- r^2 m / b) + b
= - r^2 m^2 / b + b^2 / b
= ( b^2 - r^2 m^2 ) / b
use **equation 4:r^2 = b^2 - r^2 m^2
= ( r^2 ) / b
y = r^2 / b
***part b done
x = - r^2 m / b,y = r^2 / b
the point of tangency is (-r^2m/b,r^2/b)
c) slope of line R from center of circle (0,0) to point in part B is
(x - 0) / (y - 0) = - mr^2/b / (r^2/b)
= - mr^2/b * (b/r^2) = -m
** the slope of this line is -m,this line is perpendicular to the tangent line ( with equation y = mx + b and slope m)..
part c done
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