Question

The maximum distance of the point $(4,4)$ from the circle ${{x}^{2}}+{{y}^{2}}-2x-15=0$ is
A.10
B. 9
C. 5
D. None of these

Answer 1

Check the point $(4,4)$ whether it lies on the circle, outside the circle, or inside the circle. Put $(4,4)$ in the equation of the circle ${{x}^{2}}+{{y}^{2}}-2x-15=0$ . After putting $(4,4)$ in the equation ${{x}^{2}}+{{y}^{2}}-2x-15$ we get a value. If it is greater than zero then the given point lies outside the circle and if a point lies outside a circle then its maximum distance is the summation of radius and the distance between the point $(4,4)$ and center of the circle ${{x}^{2}}+{{y}^{2}}-2x-15=0$ . We know the formula to find the radius, $\text{Radius=}\sqrt{{{\left( \dfrac{\text{coefficients}\,\text{of}\,\text{x}}{\text{2}} \right)}^{\text{2}}}\text{+}\left( \dfrac{\text{coefficients}\,\text{of}\,\text{y}}{\text{2}} \right)\text{-}\left( \text{constant}\,\text{term} \right)}$ .

Complete step-by-step answer:

According to the question we have a circle whose equation is ${{x}^{2}}+{{y}^{2}}-2x-15=0$ and also a point whose coordinate is $(4,4)$ .

In the equation of a circle, the coordinate center of the circle is given by dividing the coefficients of x and y by -2.

Here, the coefficients of x and y are -2 and 0 respectively.

Center = $\left( \dfrac{-2}{-2},\dfrac{0}{-2} \right)=(1,0)$ ……………………..(1)

In the equation of a circle, the coordinate center of the circle is given by,

$\text{Radius=}\sqrt{{{\left( \dfrac{\text{coefficients}\,\text{of}\,\text{x}}{\text{2}} \right)}^{\text{2}}}\text{+}\left( \dfrac{\text{coefficients}\,\text{of}\,\text{y}}{\text{2}} \right)\text{-}\left( \text{constant}\,\text{term} \right)}$ ……………(2)

Here, the coefficients of x and y are -2 and 0 respectively and the constant term is -15.

Now, putting these values in equation (2), we get

$\text{Radius=}\sqrt{{{\left( \dfrac{-2}{\text{2}} \right)}^{\text{2}}}\text{+}\left( \dfrac{0}{\text{2}} \right)\text{-}\left( -15 \right)}=\sqrt{1+15}=\sqrt{16}=4$ ……………………(3)

We have to find the maximum distance of the point $(4,4)$ from the circle.

For that, we need to get the exact location of the point whether it lies on the circle, outside the circle, or inside the circle.

Putting the coordinate of the point $(4,4)$ in the equation of the circle ${{x}^{2}}+{{y}^{2}}-2x-15$ , we get

& {{x}^{2}}+{{y}^{2}}-2x-15 \\\ & ={{4}^{2}}+{{4}^{2}}-2(4)-15 \\\ & =32-23 \\\ & =9 \\\ \end{aligned}$$ After putting the coordinate of the point $$(4,4)$$ in the equation of the circle $${{x}^{2}}+{{y}^{2}}-2x-15$$ , we get 9 as its value which is greater than 0. It means that the point $$(4,4)$$ lies outside the circle.  We have to find the maximum distance of the point $$(4,4)$$ from the circle. From the figure, we can see that the maximum distance of the point $$(4,4)$$ from the circle is AB which is the summation of the distance OB and radius of the circle. Using distance formula, $$OB=\sqrt{{{\left( 1-4 \right)}^{2}}+{{\left( 0-4 \right)}^{2}}}=\sqrt{{{3}^{2}}+{{4}^{2}}}=\sqrt{9+16}=\sqrt{25}=5$$ . Maximum distance = OB + radius = 5 + radius. From equation (3), we have the length of the radius which is 4 units. Maximum distance = 5 + 4 = 9. **Hence, the correct option is B.** **Note:** In this question, one might miss the negative sign in the process to find the centre of the circle. Here, by mistake one can find the center as $$\left( \dfrac{-2}{2},\dfrac{0}{2} \right)=(-1,0)$$ , which is wrong. One can also forget to add the length of the radius to the distance of the point and centre of the circle. So, keep in mind that if a point lies outside the circle then its maximum distance is the summation of the radius and the distance between the centre and the point.