
The line \[3x + 4y - k = 0\] is tangent to the circle \[{x^2} + {y^2} = 16\]. What are the possible values of \[k\]?
Answer
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Hint: We will find the centre and radius of the given circle \[{x^2} + {y^2} = 16\]. As we know, the point where the line and the circle intersect is perpendicular to the radius. Therefore, we will use the formula to find the distance of a point from a line. we will find the distance of the centre of the circle from the given line \[3x + 4y - k = 0\] and we will equate the distance to the radius. On simplification we will find the result.
Complete step by step answer:
For any general circle \[{x^2} + {y^2} + 2gx + 2fy + c = 0\], the coordinate of centre is \[( - g, - f)\] and radius is \[\sqrt {{g^2} + {f^2} - c} \].
Given circle is \[{x^2} + {y^2} = 16\]. For this, \[g = 0\] and \[f = 0\]. Therefore, centre is \[C(0,0)\] and radius is \[\sqrt {{{\left( 0 \right)}^2} + {{(0)}^2} - \left( { - 16} \right)} \] i.e., \[4\].
As we know, the point where the line and the circle intersect is perpendicular to the radius.
The perpendicular distance between the line and the centre of the circle is equal to the radius.
The perpendicular distance (\[d\]) of a point \[({x_1},{y_1})\] from a line \[ax + by + c = 0\] is given by \[d = \left| {\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|\].
In this question, we have \[d\] as radius of the circle, \[({x_1},{y_1})\] is centre of the circle \[(0,0)\] and the line is \[3x + 4y - k = 0\]. Putting all these values we get
\[ \Rightarrow 4 = \left| {\dfrac{{3 \times \left( 0 \right) + 4 \times \left( 0 \right) - k}}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2}} }}} \right|\]
On simplification, we get
\[ \Rightarrow 4 = \left| {\dfrac{{ - k}}{{\sqrt {9 + 16} }}} \right|\]
On further simplification, we get
\[ \Rightarrow 4 = \left| {\dfrac{{ - k}}{{\sqrt {25} }}} \right|\]
\[ \Rightarrow 4 = \left| {\dfrac{{ - k}}{5}} \right|\]
Using the definition of modulus function i.e., \[\left| a \right| = \pm a\], we get
\[ \Rightarrow \dfrac{{ - k}}{5} = \pm 4\]
\[ \Rightarrow \dfrac{{ - k}}{5} = 4\] or \[\dfrac{{ - k}}{5} = - 4\]
On cross multiplication, we get
\[ \Rightarrow - k = 20\] or \[ - k = - 20\]
On simplification, we get
\[ \Rightarrow k = - 20\] or \[k = 20\]
Therefore, the possible values of \[k\] is \[20\] or \[ - 20\].
Note:
A tangent intersects the circle exactly in one single point and can never cross the circle or enter it. The point where a tangent touches the circle is known as the point of tangency. Also, from one external point only two tangents can be drawn to a circle that have equal tangent segments.
Complete step by step answer:
For any general circle \[{x^2} + {y^2} + 2gx + 2fy + c = 0\], the coordinate of centre is \[( - g, - f)\] and radius is \[\sqrt {{g^2} + {f^2} - c} \].
Given circle is \[{x^2} + {y^2} = 16\]. For this, \[g = 0\] and \[f = 0\]. Therefore, centre is \[C(0,0)\] and radius is \[\sqrt {{{\left( 0 \right)}^2} + {{(0)}^2} - \left( { - 16} \right)} \] i.e., \[4\].
As we know, the point where the line and the circle intersect is perpendicular to the radius.
The perpendicular distance between the line and the centre of the circle is equal to the radius.
The perpendicular distance (\[d\]) of a point \[({x_1},{y_1})\] from a line \[ax + by + c = 0\] is given by \[d = \left| {\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|\].
In this question, we have \[d\] as radius of the circle, \[({x_1},{y_1})\] is centre of the circle \[(0,0)\] and the line is \[3x + 4y - k = 0\]. Putting all these values we get
\[ \Rightarrow 4 = \left| {\dfrac{{3 \times \left( 0 \right) + 4 \times \left( 0 \right) - k}}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2}} }}} \right|\]
On simplification, we get
\[ \Rightarrow 4 = \left| {\dfrac{{ - k}}{{\sqrt {9 + 16} }}} \right|\]
On further simplification, we get
\[ \Rightarrow 4 = \left| {\dfrac{{ - k}}{{\sqrt {25} }}} \right|\]
\[ \Rightarrow 4 = \left| {\dfrac{{ - k}}{5}} \right|\]
Using the definition of modulus function i.e., \[\left| a \right| = \pm a\], we get
\[ \Rightarrow \dfrac{{ - k}}{5} = \pm 4\]
\[ \Rightarrow \dfrac{{ - k}}{5} = 4\] or \[\dfrac{{ - k}}{5} = - 4\]
On cross multiplication, we get
\[ \Rightarrow - k = 20\] or \[ - k = - 20\]
On simplification, we get
\[ \Rightarrow k = - 20\] or \[k = 20\]
Therefore, the possible values of \[k\] is \[20\] or \[ - 20\].
Note:
A tangent intersects the circle exactly in one single point and can never cross the circle or enter it. The point where a tangent touches the circle is known as the point of tangency. Also, from one external point only two tangents can be drawn to a circle that have equal tangent segments.
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