Question

If the center of a circle $(2,3)$ and a tangent is $x+y=1$ then the equation of the circle is

Answer 1

Hint :
In the above question, we will assume that the circle's centre is a point \[\left( {{x_1},{y_1}} \right)\] because this point is on the line \[x + y = 1\] so we get an equation between \[{x_1}\] and \[{y_1}\] by replacing \[\left( {{x_1},{y_1}} \right)\] in the equation of line. We also know that the distance between the centre of a circle and any point on it is constant. So, we'll apply the idea to determine \[{x_1}\] and \[{y_1}\].

Formula used:
Circle equation is
\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]

Complete Step by step solution:
We have been provided in the question that,
The centre of a circle is
\[\left( {2,3} \right)\]
The tangent of the circle is
\[x + y = 1\]
That is equivalent to
\[y = - x + 1\].
Thus we can conclude that the Gradient of tangent is
\[ - 1\]
Radius and tangent are known to intersect at a right angle.
As a result, the radius has a gradient of \[ - 1/ - 1 = 1\] at the point where it meets this tangent.
So, \[y - 3 = 1(x - 2)\] is the equation for this radius.
On solving for y, we obtain that
y = x+1 —--- (1)
The radius of a circle is the length of the straight line that connects the centre to any point on its circumference. Because a circle's circumference can contain an endless number of points, a circle can have more than one radius.
Find the location where the radius and tangent meet:
\[x + y = 1\].
On substituting the value of y from equation (1), we get
\[x + x + 1 = 1\]
Now, we have to combine like terms, we get
\[2x + 1 = 1\]
Now, we have to subtract 1 from either side of the equation, we get
\[2x + 1 - 1 = 1 - 1\]
On simplifying we get
\[2x = 0\]
On solving for, we get
\[x = 0\]
The solutions are as follows:
\[x = 0,y = 1.\]
So, \[(0,1)\] lies on the circumference of the circle.
Consequently, the circle's radius is:
By radius formula:
We have to substitute the x and y values in the formula, we have
\[r = \sqrt {{{(2 - 0)}^2} + {{(3 - 1)}^2}} = \sqrt 8 \]
Consequently, the circle's equation is:
\[\underline {{{(x - 2)}^2} + {{(y - 3)}^2} = 8} {\rm{. }}\]
Therefore, the equation of circle is \[\underline {{{(x - 2)}^2} + {{(y - 3)}^2} = 8} {\rm{. }}\]
Note:
The general equation of a circle is another name for the centre of the circle formula. If the radius is r, the centre's coordinates are \[(h,k),\]and any point on the circle is\[(x,y)\], the centre of the circle formula is as follows:
\[{(x - h)^2} + {(y - k)^2} = {r^2}\]
The equation for the centre of the circle is another name for this. In the sections that follow, we'll use this formula to determine a circle's equation or centre.