Question 624448
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Hi,
Note: Standard Form of an Equation of a Circle is {{{(x-h)^2 + (y-k)^2 = r^2}}} 
where Pt(h,k) is the center and r is the radius

Write an equation of the line tangent to the given circle at the given point.
{{{ x^2+y^2=52 }}} (-4,6)  || Center of the circle is (0,0)
(-4,6)
(0,0) m = 6/-4 Green Line has slope m = -3/2, 
the Line Tangent at (-4,6) would be perpendicular to the Green with m = 2/3
 y = (2/3)x + b
 6 = -8/3 + b,  
  26/3 = b ,     y = (2/3)x + 26/3
{{{drawing(300,300,    -10,10,-10,10,    grid(1),
circle(0, 0,sqrt(52)),
circle(-4, 6,0.3),
circle(0, 0,0.3),

graph( 300, 300,   -10,10,-10,10, 0,-1.5x,(2/3)x + 26/3 ))}}}