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Question: How do you write an equation of a circle with center: (6 , 7) point on the circle (13 , 4)?...

How do you write an equation of a circle with center: (6 , 7) point on the circle (13 , 4)?

Explanation

Solution

Hint : Here in this question we have to find the equation of a circle where the point and center is given. By using the formula x2+y2=r2{x^2} + {y^2} = {r^2} and (x−h)2+(y−k)2=r2{(x - h)^2} + {(y - k)^2} = {r^2} , substituting the values we get the equation of a circle for the given points.

Complete step-by-step answer :
The equation of a circle can be written in the three different forms and they are:
1. standard form, where its centre is origin
x2+y2=r2{x^2} + {y^2} = {r^2}
Where x and y values of the point and the r is the radius of the circle.
2. Central form of equation of a circle
(x−h)2+(y−k)2=r2{(x - h)^2} + {(y - k)^2} = {r^2}
Where x and y values of the point, (h, k) represents the value of centre and r represents the value of radius.
3. Diameter form of equation of circle
(x−x1)(x−x2)+(y−y1)(y−y2)=0(x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0
Where x and y are the values of the point and x1,x2,y1 and y2{x_1},{x_2},{y_1}\,and\,{y_2} are the values of the end points.
Now consider the given question, here the centre is (6, 7) and the value of point is (13, 4). Here they have not given the value of radius. So first we consider the value of point and by using the standard form of an equation of a circle we determine the value of radius.
consider the standard form of equation of a circle, we have
x2+y2=r2{x^2} + {y^2} = {r^2}
Substitute the value of x and y we get
⇒132+42=r2\Rightarrow {13^2} + {4^2} = {r^2}
On squaring the terms in the LHS we get
⇒169+16=r2\Rightarrow 169 + 16 = {r^2}
On simplifying we get
⇒185=r2\Rightarrow 185 = {r^2}
Therefore we have r=185r = \sqrt {185}
Now we will consider the value of the centre (6, 7), we use the central form of the equation of the circle and then we determine the equation of a circle.
The central form is given by
(x−h)2+(y−k)2=r2{(x - h)^2} + {(y - k)^2} = {r^2}
Here the value of h is 6 , the value of y is 7 and the value of r is 185\sqrt {185}
Therefore we have (x−6)2+(y−7)2=185{(x - 6)^2} + {(y - 7)^2} = 185
On simplifying we have
⇒x2+36−12x+y2+49−14y=185\Rightarrow {x^2} + 36 - 12x + {y^2} + 49 - 14y = 185
⇒x2+y2−12x−14y+85=185\Rightarrow {x^2} + {y^2} - 12x - 14y + 85 = 185
Take 85 to the RHS we have
⇒x2+y2−12x−14y=100\Rightarrow {x^2} + {y^2} - 12x - 14y = 100
Therefore the equation of circle is given as x2+y2−12x−14y−100=0{x^2} + {y^2} - 12x - 14y - 100 = 0
So, the correct answer is “ x2+y2−12x−14y−100=0{x^2} + {y^2} - 12x - 14y - 100 = 0 ”.

Note : When the radius of the circle is not given only the value of point is given. We know that the point will lie on the circle we take the centre to origin and then by the standard form of equation of a circle we determine the value of the radius and then by the central form we write the equation of the circle.