+0  
 
0
632
4
avatar+844 

The line with equation 2x+y-5=0 is a tangent to the circle with equation 

(x-3)^2+(y-p)^2=5

a. find the two possible values of p

b.write doen the coordinates of the centre of the circle in each case.

 Nov 16, 2018
 #1
avatar+106515 
+2

2x + y - 5 = 0

(x-3)^2 + (y - p)^2  = 5

The center of the circle is  (3, p)

And the distance from this center to the point of tangency  is  √5  units

 

Using  the equation for the distance from a point to a line, we have

 

l  2(3) + 1(p) - 5 l

______________   =    √5

√[ 2^2 + 1^2 ]

 

 

l 6 + p - 5 l      =   √5

_________                     multiply both sides by √5

√5

 

 

l  6 + p  -  5 l   = 5

 

l 1 + p l   = 5

 

And we have two equations

 

1 + p  =  5             and        1 +  p   =   - 5

p = 4                                     p = -6

 

 

So.....the  coordinates are   (3, 4)    or  (3, -6)

 

See the graph here : https://www.desmos.com/calculator/asrrgimxfb

 

 

cool cool cool

 Nov 16, 2018
edited by CPhill  Nov 17, 2018
 #2
avatar+844 
0

thank you very much! may i ask where u got √[ 2^2 + 1^2 ] from?

YEEEEEET  Nov 16, 2018
 #3
avatar+2345 
+1

The formula for the distance from a point to a line is \(d=\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}\) when a line is written in the form \(Ax+By+C=0.\)

Cphill used the given equation for the line \(2x+y-5=0\). Therefore, \(A=2, B=1, \text{ and }C=-5\)

TheXSquaredFactor  Nov 17, 2018
 #4
avatar+106515 
+1

The distance from a point ( m, n)   to  line in the form  Ax + By + C  = 0  is given by :

 

l A(m) + B(n)  + C l

________________

√ [ A^2 + B^2]

 

So....  A = 2,   B = 1  and C = - 5

 

 

cool cool cool

CPhill  Nov 17, 2018

32 Online Users

avatar