The center of a circle is the point C(2,4), and P(-1,6) is a point on the circle. Find the equation of the line tangent to the circle at the point P.

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The center of a circle is the point C(2,4), and P(-1,6) is a point on the circle. Find the equation of the line tangent to the circle at the point P.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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use distanace formula to find radius
Cnu help me
D from C to P = \[\sqrt{3^2+2^2} = \sqrt{13}\] \[(x-2)^2 + (y-4)^2 = 13\]

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ok is this the answer
Yes :D
i think we write in in slope intecept form
so what would that be
You can only write a line in slope-int form, not a circle :D
ok
thanks can u help me with this one
sketch the triangle with vertics A(-5,12), B(4,-2) and C(1,-6) Find the slope of the side AC. Find the slope of the altitude form point B to side AC fIND an equation for the line that includes the altitude from point B to side AC.

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