Write the equation of a circle with centre (3,-1) which is tangent to the line y=x+2?
2 Answers
Explanation:
The equation of a circle centred at the point
Where
Hence, this question asks us to find
We know that the line
Note that this line has a slope of
We also know that the slope of any circle centred at
Since the tangent touches the circle at a point where the circle has a slope of 1, we can replace
Now, since
Finally, we can arrive at
Hence, the equation of our circle is:
This is depicted graphically below.
graph{((x-3)^2+(y+1)^2-18)(x+2-y)=0 [-18.34, 17.68, -6.8, 11.23]}
#(x-3)^2+(y+1)^2=18#
Explanation:
Given -
Centre of the circle
Equation of the tangent line -
We have to find the radius of the circle.
Find the equation of the line passing through point of tangency and the centre. It cuts the tangent at 90 degrees.
The slope of the tangent is
The slope of the required line is
Fond the equation of the required line
#mx+c=y#
#-1(3)+c=-1#
#-3+c=-1#
#c=-1+3=2#
#c=2#
#y=-x+2# ---------------(2)
Find the point where both the lines cut each other.
#y=x+2# --------------- (1)
#y=-x+2# ---------------(2) add both
#2y=4#
#y=4/2=2#
Substitute
#2=x+2#
#x=2-2=0#
The two lines cut at
Distance between
#r^2=(x_1-x_2)^2+(y_1-y_2)^2#
#r^2=(0-3)^2+(2-(-1))^2#
#r^2=(-3)^2+(2+1)^2=9+9=18#
We have Centre and radius squared
The equation is -
#(x-h)^2+(y-k)^2=r^2#
#(x-3)^2+(y+1)^2=18#