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Question: How do you write an equation with center is (4,-1) and solution point is (1, 4)?...

How do you write an equation with center is (4,-1) and solution point is (1, 4)?

Explanation

Solution

Use the circle equation (x−h)2+(y−k)2=r2{(x - h)^2} + {(y - k)^2} = {r^2}. In order to get the equation of circle, this question is asking for. The first step we need to do is to find the value of r. This can be done by using the general equation of circle, (x−h)2+(y−k)2=r2{(x - h)^2} + {(y - k)^2} = {r^2}. Here, we will put the values of (x,y) as (1,4) and (h,k) as (4,-1). From this, we will get the value of r as 6. Now, we will again use this equation to form the general equation of the required circle. Which when solved gives us the result as (x−4)2+(y+1)2=36{(x - 4)^2} + {(y + 1)^2} = 36.

Complete step by step answer:
The question has basically asked us to derive the equation of a circle with the given data
Now, we know that the general equation of a circle is given by:-
(x−h)2+(y−k)2=r2{(x - h)^2} + {(y - k)^2} = {r^2}
Where, r is the radius of the circle and x and y are the points on the circle.
Here, we have the point on the circle, but we don’t have the value of r.So, to find that, we will put the values that satisfies the circle equation that is (1,4) in place of (x,y) and (4,-1) in place of (h,k).Hence, our equation becomes:-

(x−h)2+(y−k)2=r2 (1−4)2+(4−(−1))2=r2 ⇒(−3)2+(5)2=r2 ⇒9+25=r2 ⇒r2=36 ⇒r=36 ⇒r=6 {(x - h)^2} + {(y - k)^2} = {r^2} \\\ {(1 - 4)^2} + {(4 - ( - 1))^2} = {r^2} \\\ \Rightarrow {( - 3)^2} + {(5)^2} = {r^2} \\\ \Rightarrow 9 + 25 = {r^2} \\\ \Rightarrow {r^2} = 36 \\\ \Rightarrow r = \sqrt {36} \\\ \Rightarrow r = 6 \\\

So, we have the values of the points of the circle that are (h,k) and r too.
Putting the respective values in the general circle equation, we will get

(x−4)2+(y−(−1))2=62 ∴(x−4)2+(y+1)2=36 {(x - 4)^2} + {(y - ( - 1))^2} = {6^2} \\\ \therefore {(x - 4)^2} + {(y + 1)^2} = 36 \\\

Therefore, (x−4)2+(y+1)2=36{(x - 4)^2} + {(y + 1)^2} = 36 is the required equation.

Note: The values must be correctly put when you are using the general equation.Like, the center here was (4,-1). So, we must write the second part of LHS as(y−(−1))2{(y - ( - 1))^2}and not as (y−1)2{(y - 1)^2}. Doing the later one will obviously get you to wrong answers. Therefore, signs must be used as it is given in the question always.