Question

What is the value of a such that the power of the point $(1,6)$ with respect to the circle ${{x}^{2}}+{{y}^{2}}+4x-6y-a=0$ is $-6$?
(A) $13$
(B) $11$
(C) $7$
(D) $21$

Answer 1

In the above question, we have to find the value of ‘a’ when the points on the circle are given to us. The power of the point means that the value which comes out when we put the given points in the equation. So in this question, the power of the point is given in the question.

Complete step by step answer:
A circle is a two-dimensional figure in space. In mathematics, a circle is defined as the locus of all points which are equidistant from the center of the circle. The constant distance from the center of the circle to its outer surface is known as the radius of the circle. If we double the radius of the circle then the diameter of the circle will be obtained.
The general equation of the circle is given as shown below.
${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$
The radius of the circle with the above equation is given as shown below.
$r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$
If ‘h’ and ‘k’ are the centers of the circle then,

& g=-h \\\ & f=-k \\\ \end{aligned}$$ If $${{g}^{2}}+{{f}^{2}} > c$$, then the radius of the circle comes out to be real. If $${{g}^{2}}+{{f}^{2}}=c$$, then the radius of the circle comes out to be zero and it also proves that the circle coincides with the centre of the circle and hence circle is called a point circle. If $${{g}^{2}}+{{f}^{2}} < c $$, then the radius of the circle becomes imaginary. The standard equation of the circle is given as follows $${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$$ Where ‘h’ and ‘k’ are the points of the center of the circle and ‘r’ is the radius of the circle. so, if we know the coordinates of the center of the circle and the radius of the circle then we can find out the equation of the circle. In the above question, we have to find the value of ‘a’ in the equation given below. $${{x}^{2}}+{{y}^{2}}+4x-6y-a=0$$ The value of the point ‘x’ and ‘y’ in the question is given as $$(1,6)$$ and after putting these values in the equation we get the result $$-6$$. $$\begin{aligned} & {{(1)}^{2}}+{{(6)}^{2}}+4(1)-6(6)-a=-6 \\\ & \Rightarrow 1+36+4-36-a=-6 \\\ & \Rightarrow 5-a=-6 \\\ & \Rightarrow a=11 \\\ \end{aligned}$$ So the value ‘a’ comes out to be $$11$$. **So, the correct answer is “Option B”.** **Note:** The line segment that joins one point of the circle to another is known as the chord of the circle. The largest chord of the circle is known as the diameter of the circle. The line that passes after touching one point of the circle is known as the tangent of the circle. Tangent does not pass from inside the circle.