Question

How do you graph${{x}^{2}}+{{y}^{2}}-8x+6y+16=0$?

Answer 1

To plot the graph for given equation, first of all we should know the circle equation i.e. ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$ where r is the radius and (a , b) is the centre of the circle. By converting the given equation in the form of the above equation we can easily find the centre and radius for the given equation. Therefore we can plot the graph easily.

Complete step by step answer:
From the given question, we are given to draw a graph for the given equation.
Let us consider
${{x}^{2}}+{{y}^{2}}-8x+6y+16=0.....\left( 1 \right)$
We know that the equation of a circle for radius r centred (a, b) can be written as
${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}........\left( f1 \right)$.
So, we have to convert equation (1) in the form of the above formula.
By comparing the above formula and equation (1). We can say that $-8x$ is in the format of $-2ax$and $+6y$ is in the format of$-2by$.

& -8x=-2ax \\\ & \Rightarrow a=4 \\\ \end{aligned}$$ $$\begin{aligned} & +6y=-2by \\\ & \Rightarrow b=-3 \\\ \end{aligned}$$ Therefore, the value of a and b are 4 and -3 respectively. Now for converting the equation (1) in the form of basic formula. We need $${{a}^{2}}$$and $${{b}^{2}}$$terms. As we know the values of a and b be $${{a}^{2}}$$and $${{b}^{2}}$$will be 16 and 9 respectively. But we have only 16 as constant in equation (1). So we have to add with 9 and subtract with 9 to the equation (1). By adding and subtracting with 9 we get, $${{x}^{2}}+{{y}^{2}}-8x+6y+16+9-9=0$$ Let us consider $${{x}^{2}}+{{y}^{2}}-8x+6y+16+9-9=0.......\left( 2 \right)$$ We can write equation (2) as $$\Rightarrow {{x}^{2}}+16-2.4x+{{y}^{2}}+9+2.3.y-9=0$$ By the formula $${{\left( a+b \right)}^{2}}={{a}^{2}}+2.a.b+{{b}^{2}}$$ By applying the above formula. We can rewrite the equation (2) as $$\Rightarrow {{\left( x-4 \right)}^{2}}+{{\left( y+3 \right)}^{2}}-9=0;$$ Let us consider $$\Rightarrow {{\left( x-4 \right)}^{2}}+{{\left( y+3 \right)}^{2}}-9=0........\left( 3 \right)$$ Transferring 9 from LHS (left hand side) to RHS (right hand side) in equation (3) $$\Rightarrow {{\left( x-4 \right)}^{2}}+{{\left( y+3 \right)}^{2}}=9$$ $$\Rightarrow {{\left( x-4 \right)}^{2}}+{{\left( y+3 \right)}^{2}}={{3}^{2}}$$ Let us consider $$\Rightarrow {{\left( x-4 \right)}^{2}}+{{\left( y+3 \right)}^{2}}={{3}^{2}}.....\left( 4 \right)$$ By comparing the equation (4) with equation (f1). We can easily see that equation (5) is in the form of a circle equation with radius 3 and centre (-4, 3). Therefore, graph for the equation (4) is  **Note:** Students should be well aware of plotting circle graphs. Questionnaire may ask to draw a graph for parabola also. So, students should aware of all concepts in 3D and 2D geometry. Students should also avoid calculation mistakes while solving this problem.