Question

A tangent drawn from the point $\left( {4,0} \right)$ to the circle ${x^2} + {y^2} = 8$ touches it at a point A in the first quadrant. Find the coordinates of another point B on the circle such that AB= 4.

Answer 1

Hint: In order to find the coordinate, first we will answer the coordinate of A as $\left( {h,k} \right)$ and B as $\left( {a,b} \right)$ Then we will find the tangent equation to the circle ${x^2} + {y^2} = {a^2}$ at $\left( {x,y} \right)$ which can be given as $x{x_1} + y{y_1} = {a^2},$ and by using the property we will proceed further that is if point lies on the circle then it will satisfy its equation.

Complete Step-by-Step solution:

Equation of given circle is ${x^2} + {y^2} = 8..............\left( 1 \right)$
Let A $\left( {h,k} \right)$ be the point of contact in the first quadrant of tangent from P $\left( {4,0} \right)$ to the circle (1)
Equation of tangent at A $\left( {h,k} \right)$ is $hx + ky = 8.$
It passes through P $\left( {4,0} \right)$
$
  \therefore 4h = 8 \\
   \Rightarrow h = 2 \\
 $
Since, A $\left( {h,k} \right)$ lies on the circle, we get
\[
   \Rightarrow {h^2} + {k^2} = 8 \\
   \Rightarrow 4 + {k^2} = 0 \\
   \Rightarrow k = 2{\text{ }}\left[ {\because k > 0} \right] \\
   \Rightarrow A \equiv \left( {2,2} \right) \\
 \]
$\left( {a,b} \right)$Solving (2) and (3), we get $a = 2,b = - 2{\text{ or }}a = - 2,b = 2$
Hence the coordinates of B are $\left( {2, - 2} \right)$ or $\left( { - 2,2} \right).$

Note: To solve problems related to finding the tangent of a circle, remember the equation of the circle as well as the equation of the tangent of the circle. Tangent to any curve can be found by first differentiating the curve with respect to x variables if the curve is a function of f(x), and then substituting the point at which we want to find the tangent and solving this we will get the slope. Using the equation straight line, we will get the equation of tangent.