Question

The number of feet of normal from the point $\left( {7, - 4} \right)$ to the circle ${x^2} + {y^2} = 5$ is:
A) 1
B) 2
C) 3
D) 4

Answer 1

Hint:
First, we have to check if the point $\left( {7, - 4} \right)$ lies inside the circle ${x^2} + {y^2} = 5$ or outside the circle. For that we will put the coordinates of the point in the equation of circle, and if the value comes out to be:
1) Greater than 5, then it is an exterior point.
2) Equal to 5, then point lies on the circle
3) Less than 5, then it is an interior point.
Note that the normal drawn from an external point always passes through the centre of the circle and therefore, intersects the circle at two points at the circumference.

Complete step by step solution:
First, we have to check if the point $\left( {7, - 4} \right)$ lies inside the circle ${x^2} + {y^2} = 5$ or outside it.

For this we will substitute \[x = 7\] and \[y = - 4\] in the equation of circle i.e. ${x^2} + {y^2}$.
After putting these coordinates, we get:
$
  {\left( 7 \right)^2} + {\left( { - 4} \right)^2} \\
   = 49 + 16 \\
   = 65 \\
 $
Since, 65 is greater in value than 5, therefore the point lies outside the circle.
We know that normal drawn from an external point passes through the centre of the circle and hence, intersects the circle at two points.
 Since, $\left( {7, - 4} \right)$ is an external point to the circle ${x^2} + {y^2} = 5$ there will be two feet of normal.

Hence the correct option is (B).

Note:
The normal to a curve at a given point is the line drawn perpendicular to the tangent at that point of contact. In other words, the line drawn perpendicular to the tangent (of a curve) and passing through the point of contact, is called the normal to that curve. Also, note that normal drawn from an external point passes through the centre of the circle and hence, intersects the circle at two points.