+845 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.
+129852 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
+2446 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, \(A=2, B=1, \text{ and }C=-5\)
