A circle touches two perpendicular lines #2x-3y=15# and #3x+2y=3# at the points #A(6,-1)#, #B(1,0)# respectively. Find the equation of the circle?

The tangent to a circle

#x^2+y^2+ 2ax+2by+c = 0#

at the point #(x_1,y_1)# is given by

#x x_1+yy_1+a(x+x_1)+b(y+y_1)+c = 0#

Thus, the tangent at #A = (6,-1)# is

#6x-y+a(x+6)+b(y-1)+c = 0 #

or

#(6+a)x+(b-1)y+(6a-b+c) = 0#

Since this is the line #2x-3y = 15#, we must have

#(6+a)/2 = (b-1)/(-3) = (6a+b-c)/(-15)#

Again, the tangent at #B = (1,0)# is

#1x+0times y+a(x+1)+b(y+0)+c = 0 #

or

#(1+a)x+by+(a+c) = 0#

Since this is the line #3x+2y = 3#, we must have

#(1+a)/3 = (b)/(2) = (a+c)/(-3)#

Hence #b = 2/3(1+a)# and substituting this in #(6+a)/2 = (b-1)/(-3) # gives

#(6+a)/2 = (2/3(1+a)-1)/(-3) = -2/9 a+1/9 implies (1/2+2/9)a = 1/9-3#

Thus

#13/18a = -26/9 implies a= -4#

So,

#b = 2/3(a+1) = -2#

and

# (b)/(2) = (a+c)/(-3) implies -4+c = -3/2times (-2) = 3 implies c = 7#

Thus the circle is

#x^2+y^2-8x-4y+7=0#

The following is a graph of:

#color(red)(2x-3y=15)" [1]"#

#color(blue)(3x+2y=3)" [2]"#

#color(green)(A(6,-1))#, and #color(orange)(B(1,0))#

Because the center must be located on lines that are perpendicular to the points of tangency, we can use equations [1] and [2] to write two equations that must intersect at the center.

Set equation [1] equal to an arbitrary constant, #c_1#:

#2x-3y= c_1" [1.1]"#

Set equation [2] equal to an arbitrary constant #c_2#:

#3x+2y=c_2" [2.1]"#

Substitute point B into equation [1.1] and solve for #c_1#

#2(1)-3(0)= c_1#

#c_1 = 2#

Substitute the above into equation [1.1]:

#2x-3y= 2" [1.2]"#

Substitute point A into equation [2.1] and solve for #c_2#:

#3(6)+2(-1)=c_2#

#c_2=16#

Substitute the above into equation [2.1]:

#3x+2y=16" [2.2]"#

Add equations [1.2] and [2.2] to the graph:

Please understand that the center of the circle must be at the intersection of equations [1.2] and [2.2], therefore, we shall solve them as a system of equations:

#2x-3y= 2" [1.2]"#
#3x+2y=16" [2.2]"#

#4x-6y= 4" [1.3]"#
#9x+6y=48" [2.2]"#

#13x=52#

#x = 4#

#3(4)+2y=16#

#2y=4#

#y = 2#

The center of the circle is at the point #C(4,2)#

The standard Cartesian equation of a circle is:

#(x-h)^2+(y-k)^2=r^2" [3]"#

where #(x,y)# is any point on the circle, #(h,k)# is the center, and #r# is the radius.

Substitute point C into equation [3]:

#(x-4)^2+(y-2)^2=r^2" [3.1]"#

To find the radius, we shall substitute point B into equation [3.1]:

#(1-4)^2+(0-2)^2=r^2#

#9+4=r^2#

#13 = r^2#

#r = sqrt13#

Substitute the above into equation [3.1]:

#(x-4)^2+(y-2)^2=(sqrt13)^2" [3.2]"#

Add equation [3.2] to the graph:

Please observe that equation [3.2] touches [1] and [2] and points A and B respectively, therefore, equation [3.2] is the correct equation of the circle.