Question

Find the equation of the circle touching the line 4x-3y=28 at (4,-4) and passing through (-3,-5)

Answer 1

The center of the circle is on the line perpendicular to the given line thru (4,-4).
The slope of the line is -3/4 --> y+4 = (-3/4)*(x-4)
--> y = -3x/4 - 1
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Answer 2

Hi friend !!

Here's ur ans :-

• The center will be on the line perpendicular to 4x - 3y = 28 at (4,-4). 
That line is 4y = -3x - 4. 

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So the center has the form (x, (-3x-4)/4). 
The squared distances from the center to the two given points must be equal, so 

(x-4)^2 + ( (-3x-4)/4 + 4 )^2 = (x+3)^2 + ( (-3x-4)/4 + 5 )^2 

Simplify and get x=0. Thus the center is at (0,-1). 


The squared radius using distance formula and the points (0,-1), (4,-4) is r^2 = 4^2 + 3^2 = 5^2. 


So the circle's equation is 

x^2 + (y+1)^2 = 25