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Question: How do you write the equation in standard form \({x^2} + {y^2} - 4x - 14y + 29 = 0\) ?...

How do you write the equation in standard form x2+y2−4x−14y+29=0{x^2} + {y^2} - 4x - 14y + 29 = 0 ?

Explanation

Solution

We have been given a quadratic equation in two variables. Such an equation represents a conic section in coordinate geometry. To write the equation in the standard form we have to first identify the type of conic section that this given equation represents. Then we can simplify the equation to convert it into standard form.

Complete step by step solution:
We have been given an equation x2+y2−4x−14y+29=0{x^2} + {y^2} - 4x - 14y + 29 = 0.
This quadratic equation in two variables is similar to the general form of equation of a conic section.
The general equation of a conic section is given as,
Ax2+Bxy+Cy2+Dx+Ey+F=0A{x^2} + Bxy + C{y^2} + Dx + Ey + F = 0
On comparing we can observe,
A=1,  B=0,  C=1,  D=−4,  E=−14,  F=29A = 1,\;B = 0,\;C = 1,\;D = - 4,\;E = - 14,\;F = 29
We first calculate the discriminant B2−4AC{B^2} - 4AC.
B2−4AC=0−4=−4{B^2} - 4AC = 0 - 4 = - 4
Since the discriminant is less than zero, the given equation represents either a circle or an ellipse.
We have another condition that if A=CA = C it is an equation of a circle, else it will be an equation of an ellipse.
Since, A=C=1A = C = 1 we can say that the given equation represents a circle.
Now the standard form of a circle is given as (x−a)2+(y−b)2=r2{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}, where (a,  b)\left( {a,\;b} \right) is the center and rr is the radius.
To convert the equation into a standard form of circle, we group the terms of both the variables separately.
x2+y2−4x−14y+29=0 ⇒(x2−4x)+(y2−14y)+29=0  {x^2} + {y^2} - 4x - 14y + 29 = 0 \\\ \Rightarrow \left( {{x^2} - 4x} \right) + \left( {{y^2} - 14y} \right) + 29 = 0 \\\
For completing the square we can write,
⇒(x2−4x+4)+(y2−14y+49)+29−53=0 ⇒(x−2)2+(y−7)2−24=0 ⇒(x−2)2+(y−7)2=(26)2  \Rightarrow \left( {{x^2} - 4x + 4} \right) + \left( {{y^2} - 14y + 49} \right) + 29 - 53 = 0 \\\ \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - 7} \right)^2} - 24 = 0 \\\ \Rightarrow {\left( {x - 2} \right)^2} + {\left( {y - 7} \right)^2} = {\left( {2\sqrt 6 } \right)^2} \\\
This is the final form of the equation in the standard form.

Note: To convert the given equation into standard form we have to first identify which type of conic section the equation represents. This is identified by comparing the equation with the general form and then using the conditions for different conic sections. Calculation errors should be taken care of while transforming the equation in standard form.