Find an equation of the circle with center at (-3,-5) and tangent to the line 12x+5y=4.?

Question

Please I need help on that.

Answer 1

Distance of point (-3,-5) from line 12x + 5y = 4 is the radius of the circle.

So, r = I-36-25-4I / sqrt(12^2 + 5^2)

r = 65/13 = 5

Equation of circle with center (x1,y1) and radius r is
(x-x1)^2 + (y-y1)^2 = r^2

so, (x+3)^2 + (y+5)^2 = 13^2

x^2 + y^2 + 6x + 10y + 34 = 169

so, equation of cirlce is

x^2 + y^2 + 6x + 10y = 135

Answer 2

DG has it goin' on right up to the point where he substitutes the wrong value for the radius back into the equation for the circle. It should be
(x+3)²+(y+5)²=5² and not 13² so that
x²+6x+y²+10y+9 = 0 is the equation of the desired circle.

I sure am glad to find out that I'm not the only one who does things like that ☺


Doug

Answer 3

DG's approach is correct but, after correctly finding the radius to be 5, he mistakenly puts r = 13 in the equation of the circle. The correct answer is x^2 + y^2 +6x +10y +9 = 0

Answer 4

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