How do you solve \[{{\left( x-3 \right)}^{2}}+{{\left( y-5 \right)}^{2}}=25\] and \[{{x}^{2}}+{{\left( y-1 \right)}^{2}}=100\] ?

Hint:In the given question, you have been asked to solve the two equations simultaneously. In order to solve the question, first we need to expand the equation by using the property of \[{{\left( a-b \right)}^{2}}\]to get the linear equation. Solve the linear equation for the value of \[x\] and \[y\]. Substitute back the value of either \[x\ or\ y\] in one of the linear equations.

Formula used:
\[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]

Complete step by step answer:
We have the given equation:
\[{{\left( x-3 \right)}^{2}}+{{\left( y-5 \right)}^{2}}=25\]----- (1)
\[\Rightarrow{{x}^{2}}+{{\left( y-1 \right)}^{2}}=100\]------- (2)
By using the property of \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\],
Expanding the square in the equation (1)
\[\left( {{x}^{2}}-6x+9 \right)+{{\left( {{y}^{2}}-10x+25 \right)}^{2}}=25\]
Simplifying the equation (1), we get
\[{{x}^{2}}-6x+9+{{y}^{2}}-10y+25=25\]
\[\Rightarrow{{x}^{2}}-6x+{{y}^{2}}-10y=-9\] ------ (3)
Expanding the square in the equation (2)
\[{{x}^{2}}+\left( {{y}^{2}}-2y+1 \right)=100\]
Simplifying the equation (2), we get
\[{{x}^{2}}+{{y}^{2}}-2y+1=100\]
\[\Rightarrow{{x}^{2}}+{{y}^{2}}-2y=99\]------ (4)

Subtracting equation (3) from (4), we get
\[6x+8y=108\]----- (5)
Simplifying the equation for the value of \[y\], we get
\[y=\dfrac{54-3x}{4}\]------ (6)
Substitute the value of \[y=\dfrac{54-3x}{4}\] in equation (2), we get
\[\Rightarrow {{x}^{2}}+{{\left( \dfrac{54-3x}{4}-1 \right)}^{2}}=100\]
\[\Rightarrow {{x}^{2}}+{{\left( \dfrac{54-3x}{4}-\dfrac{4}{4} \right)}^{2}}=100\]
\[\Rightarrow {{x}^{2}}+{{\left( \dfrac{50}{4}-\dfrac{3x}{4} \right)}^{2}}=100\]
Expanding the square, we get
\[\Rightarrow {{x}^{2}}+{{\left( \dfrac{2500}{16}-\dfrac{300x}{16}+\dfrac{9{{x}^{2}}}{16} \right)}^{2}}=100\]

Multiply both the sides of the equation by 16, we get
\[\Rightarrow 16{{x}^{2}}+2500-300x+9{{x}^{2}}=1600\]
Simplifying the above equation, we get
\[\Rightarrow 25{{x}^{2}}-300x+900\]
Write the above quadratic equation in standard form i.e. \[a{{x}^{2}}+bx+c=0\]
\[\Rightarrow 25{{x}^{2}}-300x+900=0\]
Divide both the side of the equation by 25, we get
\[\Rightarrow {{x}^{2}}-12x+36=0\]
Splitting the middle term of the equation, we get
\[\Rightarrow {{x}^{2}}-6x-6x+36=0\]
Taking out common factor by forming a pair, we get
\[\Rightarrow x\left( x-6 \right)-6\left( x-6 \right)=0\]

Combine the common factors, we get
\[\Rightarrow \left( x-6 \right)\left( x-6 \right)=0\]
\[\Rightarrow {{\left( x-6 \right)}^{2}}=0\]
\[\Rightarrow x=6\]
Substitute the value of \[x=6\] in the equation (6), we get
\[\Rightarrow y=\dfrac{54-3x}{4}\]
\[\Rightarrow y=\dfrac{54-3\times 6}{4}\]
\[\Rightarrow y=\dfrac{36}{4}=9\]
\[\therefore y=9\]

Therefore, the value of \[x=6\], \[y=9\] are the required solution.

Note:In the given question, they are given simultaneous equations because the equations need to be solved at the same time. In the solution, you get the value of x and y should satisfy both the equations, not only one question given in the question. To solve the equation, here means we need to find the values of the variables given in the equation.