Question

Find the equation of the circle passing through (4, 1) and (6, 5) and whose center is on line 4x + y = 16.

Answer 1

Hint: To solve the question, we have to apply the formula of the distance between two points and the formula of the general circle equation and have to analyze that the given points are equidistant from the center of the circle. Thus, by substituting the given points in the distance between two points formula and solving we can calculate the values of center and radius to the circle.

Complete step-by-step solution

Let (h, k) be the center of the given circle.
We know that the equation of the circle with centre (h, k) and radius r is given by \[{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}\] ………….. (1)
We know that the points which lie on the circle are equidistant from the centre of the circle with the distance equal to the radius of the circle.
Since, given that the points (4, 1) and (5, 6) lie on the circle. This implies the distance between (4, 1) and (6, 5) from the centre (h, k) are equal.
We know the formula for the distance between two points \[\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\] is given by \[\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}\]
Thus, by applying the given values we get
The distance between (4, 1) and (h, k) = \[\sqrt{{{\left( 4-h \right)}^{2}}+{{\left( 1-k \right)}^{2}}}\]
The distance between (6, 5) and (h, k) = \[\sqrt{{{\left( 6-h \right)}^{2}}+{{\left( 5-k \right)}^{2}}}\]
Since the distance are equal, we get
\[\sqrt{{{\left( 4-h \right)}^{2}}+{{\left( 1-k \right)}^{2}}}=~\sqrt{{{\left( 6-h \right)}^{2}}+{{\left( 5-k \right)}^{2}}}\]
Squaring on both sides of the equation, we get
\[\begin{align}
  & {{\left( \sqrt{{{\left( 4-h \right)}^{2}}+{{\left( 1-k \right)}^{2}}} \right)}^{2}}=~{{\left( \sqrt{{{\left( 6-h \right)}^{2}}+{{\left( 5-k \right)}^{2}}} \right)}^{2}} \\
 &\Rightarrow {{\left( 4-h \right)}^{2}}+{{\left( 1-k \right)}^{2}}={{\left( 6-h \right)}^{2}}+{{\left( 5-k \right)}^{2}} \\
\end{align}\]
We know the formula \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]
By applying the above formula to the above equation, we get
\[{{4}^{2}}+{{h}^{2}}-2(4)h+{{1}^{2}}+{{k}^{2}}-2(1)k={{6}^{2}}+{{h}^{2}}-2(6)h+{{5}^{2}}+{{k}^{2}}-2(5)k\]
\[\Rightarrow 16+{{h}^{2}}-8h+1+{{k}^{2}}-2k=36+{{h}^{2}}-12h+25+{{k}^{2}}-10k\]
By rearranging the terms and cancelling the common terms on both sides of the equation, we get
\[16-8h+1-2k=36-12h+25-10k\]
\[\begin{align}
  & 16-8h+1-2k=36-12h+25-10k \\
 &\Rightarrow 17-8h-2k=61-12h-10k \\
 &\Rightarrow 12h+10k-8h-2k=61-17 \\
\end{align}\]
$4h + 8k = 44 $ …… (2)
Since given that the centre of circle (h, k) lie on 4x + y = 16
We get
$4h + k = 16$
By subtracting the above equation from the equation (2) we get
$4h + 8k – (4h + k) = 44 – 16$
$\Rightarrow 7k = 28$
\[\Rightarrow k=\dfrac{28}{7}=4\]
By substituting the equation (2) in the above equation we get
\[\begin{align}
  & 4h+8\times 4=44 \\
 &\Rightarrow 4h+32=44 \\
 &\Rightarrow 4h=44-32 \\
 &\Rightarrow 4h=12 \\
 &\Rightarrow h=\dfrac{12}{4}=3 \\
\end{align}\]
Thus, we get the centre of the circle (h, k) is equal to (3, 4)
The radius of the circle = The distance between (4, 1) and (h, k)
By substituting the distance between two points formula, we get
\[r=\sqrt{{{\left( 4-h \right)}^{2}}+{{\left( 1-k \right)}^{2}}}\]
By substituting the values in the above equation, we get
\[\begin{align}
  & r=\sqrt{{{\left( 4-3 \right)}^{2}}+{{\left( 1-4 \right)}^{2}}} \\
 &\Rightarrow r=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( -3 \right)}^{2}}} \\
 &\Rightarrow r=\sqrt{1+9} \\
 &\Rightarrow r=\sqrt{10} \\
\end{align}\]
Thus, the radius of the circle is equal to \[\sqrt{10}\] units.
By substituting the values in the general equation of the circle (1), we get
\[\begin{align}
  & {{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}={{\left( \sqrt{10} \right)}^{2}} \\
 &\Rightarrow {{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}=10 \\
\end{align}\]
We know that \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]. Thus, by substituting the formula in the above equation, we get
\[\begin{align}
  &\Rightarrow {{x}^{2}}+{{3}^{2}}-2(3)x+{{y}^{2}}+{{4}^{2}}-2(4)y=10 \\
 &\Rightarrow {{x}^{2}}+9-6x+{{y}^{2}}+16-8y=10 \\
 &\Rightarrow {{x}^{2}}+{{y}^{2}}-6x-8y+25=10 \\
 &\Rightarrow {{x}^{2}}+{{y}^{2}}-6x-8y+25-10=0 \\
 &\Rightarrow {{x}^{2}}+{{y}^{2}}-6x-8y+15=0 \\
\end{align}\]
Thus, the equation of the given circle is equal to \[{{x}^{2}}+{{y}^{2}}-6x-8y+15=0\]

Note: The possibility of mistake can be done by not applying the formula of the distance between two points and the formula of the general circle equation. The other possibility of a mistake can be done not by analyzing that the given points are equidistant from the center of the center. The alternative way of calculating the center of a circle is by substituting the given points in the circle equation and solving those equations to calculate the value of h, k, r.