Question

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?

Answer 1

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

Explanation:

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`

Answer 2

`(x-4)^2+(y-2)^2=(sqrt13)^2`

Explanation:

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.