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${{x}^{2}}+{{y}^{2}}-14x-10y-151=0$

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Hint: Find the distance of the point from the centre of the circle. After doing so, add the radius to it for the largest distance, and subtract the radius from it for shortest distance.

Complete step-by-step answer:

As, the general equation of the circle is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ where $(h,k)$ is the centre and $r$ is the radius.

Now the given equation is : ${{x}^{2}}+{{y}^{2}}-14x-10y-151=0$

Adding $49+25$ to both sides, we get :

$\begin{align}

& \Rightarrow {{x}^{2}}-14x+49+{{y}^{2}}-10y+25=151+49+25 \\

& \Rightarrow {{\left( x-7 \right)}^{2}}+{{\left( y-5 \right)}^{2}}=225 \\

& \Rightarrow {{\left( x-7 \right)}^{2}}+{{\left( y-5 \right)}^{2}}={{\left( 15 \right)}^{2}} \\

\end{align}$

So, the circle has its centre at $\left( 7,5 \right)$ and has a radius $r=15$ units.

Then, the shortest distance between the point $({{x}_{1}},{{y}_{1}})$ and the circle is given by the distance of that point from the centre of the circle minus the radius of the circle.

Hence, if $d$ is the shortest distance of the point $({{x}_{1}},{{y}_{1}})$ from the circle, then :

\[\begin{align}

& d=\left| \sqrt{{{\left( {{x}_{1}}-h \right)}^{2}}+{{\left( {{y}_{1}}-k \right)}^{2}}}-r \right| \\

& =\left| \sqrt{{{\left( 2-7 \right)}^{2}}+{{\left( -7-5 \right)}^{2}}}-15 \right| \\

& =\left| \sqrt{{{\left( -5 \right)}^{2}}+{{\left( -12 \right)}^{2}}}-15 \right| \\

& =\left| \sqrt{169}-15 \right|=|13-15|=2 \\

\end{align}\]

Therefore, the shortest distance of point with coordinates $({{x}_{1}},{{y}_{1}})$ from the circle$=d=2\text{ Unit }$

Now, the longest distance between the point $({{x}_{1}},{{y}_{1}})$ and the circle is defined as the sum of the distance of point $({{x}_{1}},{{y}_{1}})$ from the centre of the circle, and the circle’s radius.

Hence, if $D$ is the largest distance of the point $({{x}_{1}},{{y}_{1}})$ from the circle, then :

\[\begin{align}

& D=\left| \sqrt{{{\left( {{x}_{1}}-h \right)}^{2}}+{{\left( {{y}_{1}}-k \right)}^{2}}}+r \right| \\

& =\left| \sqrt{{{\left( -5 \right)}^{2}}+{{\left( -12 \right)}^{2}}}+15 \right| \\

& =\left| \sqrt{169}+15 \right|=|13+15|=28 \\

\end{align}\]

Therefore, the longest distance of the point $({{x}_{1}},{{y}_{1}})$ from the circle $=D=28\text{ units}\text{.}$

Note: The most common mistake students make is in finding points and remembering formulae.

Remember that, \[D=\left| \sqrt{{{\left( {{x}_{1}}-h \right)}^{2}}+{{\left( {{y}_{1}}-k \right)}^{2}}}+r \right|\text{ }\]always, \[d=\left| \sqrt{{{\left( {{x}_{1}}-h \right)}^{2}}+{{\left( {{y}_{1}}-k \right)}^{2}}}-r \right|\text{ }\]always. There might also be some confusion in finding the centre point and the radius. So, you can just find the compare the equation given with the general equation of a circle which is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$, where the circle’s centre $=(-g,-f)$, and its radius $=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$. Also, be very cautious and avoid calculation mistakes.

Complete step-by-step answer:

As, the general equation of the circle is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ where $(h,k)$ is the centre and $r$ is the radius.

Now the given equation is : ${{x}^{2}}+{{y}^{2}}-14x-10y-151=0$

Adding $49+25$ to both sides, we get :

$\begin{align}

& \Rightarrow {{x}^{2}}-14x+49+{{y}^{2}}-10y+25=151+49+25 \\

& \Rightarrow {{\left( x-7 \right)}^{2}}+{{\left( y-5 \right)}^{2}}=225 \\

& \Rightarrow {{\left( x-7 \right)}^{2}}+{{\left( y-5 \right)}^{2}}={{\left( 15 \right)}^{2}} \\

\end{align}$

So, the circle has its centre at $\left( 7,5 \right)$ and has a radius $r=15$ units.

Then, the shortest distance between the point $({{x}_{1}},{{y}_{1}})$ and the circle is given by the distance of that point from the centre of the circle minus the radius of the circle.

Hence, if $d$ is the shortest distance of the point $({{x}_{1}},{{y}_{1}})$ from the circle, then :

\[\begin{align}

& d=\left| \sqrt{{{\left( {{x}_{1}}-h \right)}^{2}}+{{\left( {{y}_{1}}-k \right)}^{2}}}-r \right| \\

& =\left| \sqrt{{{\left( 2-7 \right)}^{2}}+{{\left( -7-5 \right)}^{2}}}-15 \right| \\

& =\left| \sqrt{{{\left( -5 \right)}^{2}}+{{\left( -12 \right)}^{2}}}-15 \right| \\

& =\left| \sqrt{169}-15 \right|=|13-15|=2 \\

\end{align}\]

Therefore, the shortest distance of point with coordinates $({{x}_{1}},{{y}_{1}})$ from the circle$=d=2\text{ Unit }$

Now, the longest distance between the point $({{x}_{1}},{{y}_{1}})$ and the circle is defined as the sum of the distance of point $({{x}_{1}},{{y}_{1}})$ from the centre of the circle, and the circle’s radius.

Hence, if $D$ is the largest distance of the point $({{x}_{1}},{{y}_{1}})$ from the circle, then :

\[\begin{align}

& D=\left| \sqrt{{{\left( {{x}_{1}}-h \right)}^{2}}+{{\left( {{y}_{1}}-k \right)}^{2}}}+r \right| \\

& =\left| \sqrt{{{\left( -5 \right)}^{2}}+{{\left( -12 \right)}^{2}}}+15 \right| \\

& =\left| \sqrt{169}+15 \right|=|13+15|=28 \\

\end{align}\]

Therefore, the longest distance of the point $({{x}_{1}},{{y}_{1}})$ from the circle $=D=28\text{ units}\text{.}$

Note: The most common mistake students make is in finding points and remembering formulae.

Remember that, \[D=\left| \sqrt{{{\left( {{x}_{1}}-h \right)}^{2}}+{{\left( {{y}_{1}}-k \right)}^{2}}}+r \right|\text{ }\]always, \[d=\left| \sqrt{{{\left( {{x}_{1}}-h \right)}^{2}}+{{\left( {{y}_{1}}-k \right)}^{2}}}-r \right|\text{ }\]always. There might also be some confusion in finding the centre point and the radius. So, you can just find the compare the equation given with the general equation of a circle which is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$, where the circle’s centre $=(-g,-f)$, and its radius $=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$. Also, be very cautious and avoid calculation mistakes.

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