Answer
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Hint: Start by drawing a representative diagram of the above situation using the equation of the circle. Now from the equation of the circle, find the radius and the centre of the given circle followed by finding the length of the tangent using the Pythagoras theorem.
Complete step-by-step answer:
Let us start by drawing a representative diagram of the situation mentioned in the question.
Now comparing the equation ${{x}^{2}}+{{y}^{2}}+6x-4y-3=0$ with general equation ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$, we can deduce that:
g = 3, f = -2 and c = -3.
Now,
Centre of the circle: (-g,-f)=(-3,2)
Radius of the circle= OB $=\sqrt{{{g}^{2}}+{{f}^{2}}-c}=\sqrt{9+4+3}=\sqrt{16}=4\text{ units}$
We can also find the length of OA using the distance formula as:
$OA=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
$\Rightarrow OA=\sqrt{{{\left( 5+3 \right)}^{2}}+{{\left( 1-2 \right)}^{2}}}=\sqrt{64+1}=\sqrt{65}\text{ units}$
Now from the diagram, we can say that ABO is a right-angled triangle with $\angle ABO=90{}^\circ $ . Therefore, applying Pythagoras theorem, we get
${{\left( perpendicular \right)}^{2}}+{{\left( base \right)}^{2}}={{\left( hypotenuse \right)}^{2}}$
$\Rightarrow O{{B}^{2}}+A{{B}^{2}}=O{{A}^{2}}$
Now we will substitute the values of OB and OA from above results. On doing so, we get
$\Rightarrow {{4}^{2}}+A{{B}^{2}}={{\left( \sqrt{65} \right)}^{2}}$
$\Rightarrow A{{B}^{2}}=65-16$
$\Rightarrow AB=\sqrt{49}=7\text{ units}$
Therefore, we can conclude that the length of the tangent from point (5,1) to the circle ${{x}^{2}}+{{y}^{2}}+6x-4y-3=0$ is 7 units. Hence, the answer to the above question is option (c).
Note: If you want, you can directly use the property that length of a tangent to a circle from a given point is equal to $\sqrt{{{S}_{1}}}$ , where ${{S}_{1}}$ is the value you get by putting the point from which tangent is drawn in the equation of the circle.
Complete step-by-step answer:
Let us start by drawing a representative diagram of the situation mentioned in the question.
Now comparing the equation ${{x}^{2}}+{{y}^{2}}+6x-4y-3=0$ with general equation ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$, we can deduce that:
g = 3, f = -2 and c = -3.
Now,
Centre of the circle: (-g,-f)=(-3,2)
Radius of the circle= OB $=\sqrt{{{g}^{2}}+{{f}^{2}}-c}=\sqrt{9+4+3}=\sqrt{16}=4\text{ units}$
We can also find the length of OA using the distance formula as:
$OA=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
$\Rightarrow OA=\sqrt{{{\left( 5+3 \right)}^{2}}+{{\left( 1-2 \right)}^{2}}}=\sqrt{64+1}=\sqrt{65}\text{ units}$
Now from the diagram, we can say that ABO is a right-angled triangle with $\angle ABO=90{}^\circ $ . Therefore, applying Pythagoras theorem, we get
${{\left( perpendicular \right)}^{2}}+{{\left( base \right)}^{2}}={{\left( hypotenuse \right)}^{2}}$
$\Rightarrow O{{B}^{2}}+A{{B}^{2}}=O{{A}^{2}}$
Now we will substitute the values of OB and OA from above results. On doing so, we get
$\Rightarrow {{4}^{2}}+A{{B}^{2}}={{\left( \sqrt{65} \right)}^{2}}$
$\Rightarrow A{{B}^{2}}=65-16$
$\Rightarrow AB=\sqrt{49}=7\text{ units}$
Therefore, we can conclude that the length of the tangent from point (5,1) to the circle ${{x}^{2}}+{{y}^{2}}+6x-4y-3=0$ is 7 units. Hence, the answer to the above question is option (c).
Note: If you want, you can directly use the property that length of a tangent to a circle from a given point is equal to $\sqrt{{{S}_{1}}}$ , where ${{S}_{1}}$ is the value you get by putting the point from which tangent is drawn in the equation of the circle.
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