Question

find the equation of the circle passing through the point (2,4) and centre at the point of intersection of the lines x-y=4 and 2x+3y=-7

Answer 1

Please see the attachment

Answer 2

find a circle passing through a given point and center as the point of intersection of given two lines

Explanation:

1. given two line $x-y-4=0 ,\ \ \ \ 2x+3y+7=0$  let there point of intersection be (a, b) the we have,                                                                              $a-b-4=0\\2a+3b+7=0$                        
2. solving these equations we get the point of intersection $(1,-3)$ hence, center of circle is $(1,-3)$
3. now equation of a circle with center (h, k) is given by, $(x-h)^2+(y-k)^2=r^2$  
4. we have, $h=1,\ \ k=-3$ therefore equation of circle is $(x-1)^2+(y+3)^2=r^2$
5. given that circle passes through point $(2,4)$ we get,                                                      $(2-1)^2+(4+3)^2=r^2\\r^2={1+49} \\r^2=50$
6. hence final equation of circle is , $(x-1)^2+(y+3)^2=50$