Question

The tangent to the circle \[x^{2} + y^{2} = 9\] which is parallel to the y axis and does not lie in the third quadrant touches the circle at the point
A. \[( - 3,\ 0)\]
B. \[(3,\ 0)\]
C. \[(0,\ 3)\]
D. \[(0,\ - 3)\]

Answer 1

Hint:In this question, we need to find the point which touches the circle. Also given the tangent to the circle \[x^{2} + y^{2} = 9\] which is parallel to the y axis. We need to know that the centre of the circle \[(h,k)\] is \[(0,0)\] . First, we can find the value of \[x\] from that we can easily find the value of \[y\]. By using the equation of the circle, we can find the point \[x\]. Then by substituting the value of \[x\] in the tangent given, we can find the value of \[y\].

Complete step by step answer:
Given, the tangent to the circle \[x^{2} + y^{2} = 9\] ••• (1) which is parallel to the y axis.
Any line which is parallel to y axis is
\[\Rightarrow \ x = k\] ••• (2)
Let us consider the centre of the circle, \[(h,\ k)\] as \[(0,\ 0)\]. We know that the equation of the circle is \[\left( x – h \right)^{2} - \left( y – k \right)^{2} = r^{2}\] .

Now, we can rewrite the given equation \[x^{2} + y^{2} = 9\] as \[\left( x – 0 \right)^{2} + \left( y – 0 \right)^{2} = 9\] with centre \[(0,\ 0)\]
Now on comparing both the equation,
We get ,
\[\Rightarrow \ r^{2} = 9\]
On taking square root on both sides,
We get,
\[\Rightarrow \ r = 3\]
Thus we get \[r = 3\]

If the point touches the circle \[x^{2} + y^{2} = 9\] , the perpendicular distance from the centre \[(0,0)\] to the line \[x = k\], must be equal to radius \[3\].
That is \[\dfrac{\left| 0 – k \right|}{1} = 3\]
On simplifying,
We get,
\[\Rightarrow \ k = \pm 3\]
Thus \[k = 3\] (since the line doesn’t lie in the third quadrant )
From equation (2) ,
We get conclude that \[x = 3\]
Now on substituting the value of \[x\] in the equation (1) ,
We get,
\[\Rightarrow \ \left( 3 \right)^{2} + y^{2} = 9\]

On simplifying,
We get,
\[\Rightarrow \ 9 + y^{2} = 9\]
By subtracting \[9\] on both sides,
We get,
\[\Rightarrow \ y^{2} = 9 – 9\]
On simplifying,
We get,
\[\Rightarrow \ y^{2} = 0\]
That is \[y = 0\]
Hence the point \[(x,\ y)\] is \[(3,\ 0)\] which doesn’t lie in the third quadrant.
Thus we get the point \[(x,\ y)\] which touches the circle as \[(3,\ 0)\] .

Hence, the correct answer is option B.

Note:In order to solve these types of questions, we need to have a strong grip over the equation of the circle. We may make mistakes in solving the question and also while applying the equation of the circle. So we need to be very careful in comparing the given tangent equation with the general equation of the circle. We also need to know that the tangent is nothing but a line that joins two close points from a point on the circle.