Question

find the equation of the circle for tangent to 2x – 3y – 7= 0 at (2 – 1) and passes through (4 1).

Answer 1

Equation of the circle is, $(x-\frac{2}{5} )^2+(y+\frac{13}{5} )^2$  = 128/25.

Given

To find the equation of the circle.

Let center of circle be (h,k).

Which is tangent to 2x-3y-7 = 0 at (2,-1) and it passes through (4,1).

Slope for the equation is,

                                         2x-3y-7=0

                                         y = mx + c

Comparing the above two equation,

                                        3y = 2x -7

                                          y = 2/3 x - 7/3

                      Where, m = 2/3

Its a perpendicular tangent with slope 2/3 is -3/2.

Circle passes through the point (4,1)

                      k+1 / h-2 = -3 /  2

                               3h+2k = 4  -----> (1)

Equation of circle is,

                                $(x-h)^2+(y-k)^2 = (2-h)^2+(-1-k)^2$

It passes through (4,1), (i.e.,)

                                $(4-h)^2+(1-k)^2 = (2-h)^2+(-1-k)^2$

                                   -8h+16-2k+1 = -4h+4+2k+1

                                           17-5 = 8h-4h+2k-2k

                                            4h - 4k = 12

                                            4(h-k) = 12

                                               h-k = 12/4      

                                               h-k = 3  -----> (2)

From (1) and (2), by solving

                        We get, h = 2/5   k = -13/5

                        $(x-h)^2+(y-k)^2 = (2-h)^2+(-1-k)^2$

                        $(x-\frac{2}{5} )^2+(y+\frac{13}{5} )^2 = (2-\frac{2}{5} )^2+(-1+\frac{13}{5} )^2$

                         $(x-\frac{2}{5} )^2+(y+\frac{13}{5} )^2$  = 128/25.

Therefore, equation of the circle is, $(x-\frac{2}{5} )^2+(y+\frac{13}{5} )^2$  = 128/25.

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