Step 1:

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The equation is $\"\"$ and the point is $\"\"$ .

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Consider $\"\"$ .

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Differentiate on each side with respect to $\"\"$ .

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$\"\"$

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$\"\"$

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Slope of the tangent line at $\"\"$ .

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$\"\"$

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Slope of the tangent line is $\"\"$ .

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Step 2:

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Point slope form of line equation is $\"image\"$ .

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Substitute $\"\"$ and $\"\"$ in the above equation.

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$\"\"$

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Tangent line is $\"\"$ .

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Normal line is perpendicular to tangent line then

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slope of tangent line*slope of normal line is equal to $\"\"$ .

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$\"\"$

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Point slope form of line equation is $\"image\"$ .

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Substitute $\"\"$ and $\"\"$ in the above equation.

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$\"\"$

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Normal line equation is $\"\"$ .

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Step 3:

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Equation of circle with center $\"\"$ and radius $\"\"$ is $\"\"$ .

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$\"\"$

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Differentiate on each side with respect to $\"\"$ .

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$\"\"$

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$\"\"$

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$\"\"$

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Substitute $\"\"$ .

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$\"\"$

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$\"\"$

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Substitute $\"\"$ in the above equation.

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$\"\"$

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$\"\"$

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Step 4:

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Substitute $\"\"$ in the circle equation $\"\"$

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$\"\"$

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Circle passes through the point $\"\"$ .

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$\"\"$

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$\"\"$

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Roots of the quadratic equation is $\"\"$ .

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Then,

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$\"\"$

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Therefore $\"\"$ and $\"\"$ .

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Substitute $\"\"$ in $\"\"$ .

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$\"\"$

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Substitute $\"\"$ in $\"\"$ .

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$\"\"$ .

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Step 5:

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Graph:

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Graph both the circle equations, curve and tangent line.

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$\"\"$