Question 847330
The standard form of the equation for a circle is: 

{{{ (x - h)^2 + (y + k)^2 = r^2}}} where {{{h }}} is  the {{{x}}} coordinate of the center of the circle , {{{k}}} is the {{{y}}} coordinate of the center of the circle, and {{{r }}} is the radius of the circle .

the circle that has center (4,-6) and passes through (-3,-4) 

 given:
 the center of the circle at ({{{4}}} ,{{{- 6}}}) 
one  point  the circle passes through: ({{{-3}}},{{{-4}}}) 

The only thing we are missing to complete the equation for the circle is to know (if given) or derive (if not given), {{{r}}}, the radius of the circle. 

 With the two points given: center  ({{{4}}} ,{{{- 6}}})  and circle passes through ({{{-3}}},{{{-4}}}) ; if we connect those two points together, that line will be equal the radius of the circle. 

To determine the radius, think about a right triangle that has as its "{{{a}}}" side and "{{{b}}}" side the rise and run difference between the two given points. And, knowing the Pythagorean theorem to determine the hypotenuse of a right triangle: 

{{{c^2= a^2 + b^2 }}}

 We know that in in this instance the hypotenuse is equal to the radius of our circle, so we will substitute {{{r}}} for {{{c}}}. 
{{{ r^2=a^2 + b^2 }}}

If we take the square root of both sides of the previous equation we get: 
{{{ r = sqrt(a^2 + b^2)}}}

 To solve, you need: 

{{{a}}} = the difference in the two {{{x }}}coordinates (from the two points given): 
{{{a = 4 - (-3) = 4+3=7}}} 

and {{{b }}} = the difference in the two y coordinates (from the two points given): 

{{{b =- 6 - (-4)=-6+4 = -2}}} 

then, {{{ r = sqrt(7^2 + (-2)^2)}}}

{{{ r = sqrt(49+4 ) }}}
{{{r = sqrt(53)}}} 
To return to the original problem: determine the equation for the circle with the given center point and passing through the given point: 

 {{{(x - 4)^2 + (y +6)^2 = (sqrt(53))^2}}}   which could be simplified to

 {{{ (x - 4)^2 + (y + 6)^2 =53 }}}