# Write the equation of a circle with centre (3,-1) which is tangent to the line y=x+2?

##### 2 Answers

#### Answer:

#### 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]}

#### Answer:

#(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#