SOLUTION: Find an equation for the line that is tangent to the circle x^2+y^2=169 at the point (5,12). I need help beind reminded of the equation of a circle and equations related to this,

Question 993811: Find an equation for the line that is tangent to the circle x^2+y^2=169 at the point (5,12).
I need help beind reminded of the equation of a circle and equations related to this, and also a step-by-step explantion to help me to understand how to solve this typ of problems and others efficiently. Thank you so much!
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
Center of circle on the Origin, , and for this, radius is r.
You have , so the radius .

The point ON THE CIRCLE, (5,12), is part of a tangent line which passes through this point. This means, you want to find an equation for this line and this line TOUCHES the circle at this point; and it is perpendicular to the line which contains this point (5,12) and the Origin (which is center of your circle.)
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Work to understand that discussion before continuing.

What is the line containing the circle's center (0,0) and the given point (5,12)? You should find just intuitively this is . The y-intercept and the x-intercept both 0.

What is the equation for the line PERPENDICULAR to and contains the point (5,12)? For perpendicularity, its slope must be negative reciprocal of , so this slope needed will be . You can use the point-slope equation form (for convenience if you are comfortable with it), and plug-in the needed slope and included point (5,12):

Use simple algebra if you want this equation in standard form or in slope-intercept form.

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Try to make a sketch or a graph on your own to help analyze the problem description.
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