What-is-the-equation-of-a-tangent-to-a-circle-whose-equation-is-x-2-y-2-2x-4y-1-at-point-1-5-3-

Question Number 136604 by bemath last updated on 23/Mar/21

What is the equation of a tangent to a circle whose equation is x^2+y^2−2x−4y=1 at point (1+√5,3)

Answered by EDWIN88 last updated on 23/Mar/21

step (1) is the point on the circle ?

{(1 + \sqrt{5})}^{2} + 3^{2} - 2 (1 + \sqrt{5}) - 4.3 =

6 + 2 \sqrt{5} + 9 - 2 - 2 \sqrt{5} - 12 = 1 (yes it is on the circle)

step (2) find slope the line from the center

to the given point \Rightarrow m = \frac{3 - 2}{1 + \sqrt{5} - 1} = \frac{1}{\sqrt{5}}

step (3) find the slope of tangent line

is - \frac{1}{m} = - \sqrt{5}

step (4) we get the tangent has equation

y = - \sqrt{5} (x - 1 - \sqrt{5}) + 3

y = - x \sqrt{5} + \sqrt{5} + 8

Commented by bemath last updated on 24/Mar/21

Answered by greg_ed last updated on 24/Mar/21

\boldsymbol step (1) : \boldsymbol find \boldsymbol the \boldsymbol centre \boldsymbol I \boldsymbol of \boldsymbol the \boldsymbol cercle (C) \boldsymbol given .

(C) : \boldsymbol x^{2} + \boldsymbol y^{2} - 2 \boldsymbol x - 4 \boldsymbol y = 1 \Rightarrow

\boldsymbol I (\frac{2}{2}; \frac{4}{2}) \Rightarrow

\boldsymbol I (1; 2) .

\boldsymbol step (2) : \boldsymbol M (\boldsymbol x; \boldsymbol y) \in (D) \boldsymbol the \boldsymbol tangent \boldsymbol and \boldsymbol A (1 + \sqrt{5}; 3)

\vec{\boldsymbol IA} . \vec{\boldsymbol AM} = \vec{0} \Leftrightarrow

(1 + \sqrt{5} - 1) (\boldsymbol x - 1 - \sqrt{5}) + (3 - 2) (\boldsymbol y - 3) = 0 \Leftrightarrow

\sqrt{5} (\boldsymbol x - 1 - \sqrt{5}) + \boldsymbol y - 3 = 0 \Leftrightarrow

\boldsymbol x \sqrt{5} + \boldsymbol y - \sqrt{5} - 5 - 3 = 0 \Leftrightarrow

\boldsymbol x \sqrt{5} + \boldsymbol y - \sqrt{5} - 8 = 0 \Leftrightarrow

\boldsymbol x \sqrt{5} + \boldsymbol y - (\sqrt{5} + 8) = 0 .

Commented by bemath last updated on 24/Mar/21

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