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  • A circle with radius 1 is tangent to the x-axis at the point (0,0). A line is tangent to the circle at the point (1,0). The line also passes through the point (2,3). Find the coordinates

A circle with radius 1 is tangent to the x-axis at the point (0,0). A line is tangent to the circle at the point (1,0). The line also passes through the point (2,3). Find the coordinates of the point of intersection of the line and the circle.

Option: 1

\mathrm{\left(\frac{3}{2}, \frac{3}{2}\right)}
 


Option: 2

\mathrm{\left(-\frac{3}{2}, \frac{3}{2}\right)}
 


Option: 3

\mathrm{\left(-\frac{3}{2},-\frac{3}{2}\right)}
 


Option: 4

\mathrm{\left(\frac{3}{2},-\frac{3}{2}\right)}


Answers (1)

best_answer

Equation of the circle is \mathrm{-(x-0)^2+(y-o)^2=1^2}
\mathrm{ \Rightarrow x^2+y^2=1 }
Slope of line \mathrm{=\frac{3-0}{2-1}=\frac{3}{1}}
Equation of line \mathrm{=y=\frac{3}{1} x+b}
\begin{aligned} & \Rightarrow 0=\frac{3}{1}+b \\ & \Rightarrow b=-\frac{3}{1} \end{aligned}
(since, line passes through the point \mathrm{(1,0)} )
Therefore, equation of the line is \mathrm{y=\frac{3}{1} x-\frac{3}{1}}
line and the circle intersect at the point \mathrm{(x, y)} where \mathrm{\backslash}

\mathrm{ \begin{aligned} & \Rightarrow x^2+y^2=1 \\ & y=\frac{3}{1} x-\frac{3}{1} \end{aligned} }
On solving,\mathrm{ x^2-\frac{9}{5} x+\frac{9}{10}=0}
By factorization, \mathrm{\left(x-\frac{3}{2}\right)^2=0}
So, \mathrm{x=\frac{3}{2}\: and \: y=\frac{3}{2}}
Coordinates of the point of intersection of the line and the circle are \mathrm{\left(\frac{3}{2}, \frac{3}{2}\right)}

Posted by

seema garhwal

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