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Question: Find the equation of the circle which passes through the points A (-1,4) and B (1,2) and which touch...

Find the equation of the circle which passes through the points A (-1,4) and B (1,2) and which touches the line 3x-y-3=0.

Explanation

Solution

Hint: For solving this question we will assume x2+y2−2gx−2fy+c=0{{x}^{2}}+{{y}^{2}}-2gx-2fy+c=0 to be the equation of the required circle. After that, from the given data we will form equations between gg, ff, cc and solve them to find their values.

Complete step-by-step solution -
Given:
It is given that there is a circle which passes through the points A (-1,4) and B (1,2) and which touches the line 3x−y−3=03x-y-3=0. And we have to find the equation of this circle.
Now, as we know that the general equation of the circle is x2+y2−2gx−2fy+c=0{{x}^{2}}+{{y}^{2}}-2gx-2fy+c=0 whose centre is at point (g,f)\left( g,f \right) and radius of the circle is g2+f2−c\sqrt{{{g}^{2}}+{{f}^{2}}-c} . So, let the required equation of the circle is x2+y2−2gx−2fy+c=0{{x}^{2}}+{{y}^{2}}-2gx-2fy+c=0 and as it is given that it passes through points A (-1,4) and B (1,2) so, they will satisfy the equation of the circle. Then,
x2+y2−2gx−2fy+c=0 ⇒(−1)2+(4)2−2g(−1)−2f(4)+c=0 ⇒1+16+2g−8f+c=0 ⇒2g−8f+c=−17....................(1) x2+y2−2gx−2fy+c=0 ⇒(1)2+(2)2−2g(1)−2f(2)+c=0 ⇒1+4−2g−4f+c=0 ⇒2g+4f−c=5.......................(2) \begin{aligned} & {{x}^{2}}+{{y}^{2}}-2gx-2fy+c=0 \\\ & \Rightarrow {{\left( -1 \right)}^{2}}+{{\left( 4 \right)}^{2}}-2g\left( -1 \right)-2f\left( 4 \right)+c=0 \\\ & \Rightarrow 1+16+2g-8f+c=0 \\\ & \Rightarrow 2g-8f+c=-17....................\left( 1 \right) \\\ & {{x}^{2}}+{{y}^{2}}-2gx-2fy+c=0 \\\ & \Rightarrow {{\left( 1 \right)}^{2}}+{{\left( 2 \right)}^{2}}-2g\left( 1 \right)-2f\left( 2 \right)+c=0 \\\ & \Rightarrow 1+4-2g-4f+c=0 \\\ & \Rightarrow 2g+4f-c=5.......................\left( 2 \right) \\\ \end{aligned}
Now, subtract the equation (1) from equation (2). Then,
(2g+4f−c)−(2g−8f+c)=5−(−17) ⇒4f−c+8f−c=22 ⇒12f−2c=22 ⇒2c=12f−22 ⇒c=6f−11.................................................(3) \begin{aligned} & \left( 2g+4f-c \right)-\left( 2g-8f+c \right)=5-\left( -17 \right) \\\ & \Rightarrow 4f-c+8f-c=22 \\\ & \Rightarrow 12f-2c=22 \\\ & \Rightarrow 2c=12f-22 \\\ & \Rightarrow c=6f-11.................................................\left( 3 \right) \\\ \end{aligned}
Now, add equation (1) and equation (2). Then,
(2g−8f+c)+(2g+4f−c)=−17+5 ⇒4g−4f=−12 ⇒g−f=−3 ⇒g=f−3.............................................(4) \begin{aligned} & \left( 2g-8f+c \right)+\left( 2g+4f-c \right)=-17+5 \\\ & \Rightarrow 4g-4f=-12 \\\ & \Rightarrow g-f=-3 \\\ & \Rightarrow g=f-3.............................................\left( 4 \right) \\\ \end{aligned}
Now, as it is given that the line 3x−y−3=03x-y-3=0 is tangent to the circle. Thus, when we put y=3x−3y=3x-3 in the equation of the circle. Then,
x2+y2−2gx−2fy+c=0 ⇒x2+(3x−3)2−2gx−2f(3x−3)+c=0 ⇒x2+9x2+9−18x−2gx−6fx+6f+c=0 ⇒10x2+x(−2g−6f−18)+9+6f+c=0 \begin{aligned} & {{x}^{2}}+{{y}^{2}}-2gx-2fy+c=0 \\\ & \Rightarrow {{x}^{2}}+{{\left( 3x-3 \right)}^{2}}-2gx-2f\left( 3x-3 \right)+c=0 \\\ & \Rightarrow {{x}^{2}}+9{{x}^{2}}+9-18x-2gx-6fx+6f+c=0 \\\ & \Rightarrow 10{{x}^{2}}+x\left( -2g-6f-18 \right)+9+6f+c=0 \\\ \end{aligned}
Now, put the value of c=6f−11c=6f-11 from equation (3) and value of g=f−3g=f-3 from equation (4) in the above equation. Then,
10x2+x(−2g−6f−18)+9+6f+c=0 ⇒10x2+x(−2f+6−6f−18)+9+6f+6f−11=0 ⇒10x2−x(8f+12)+12f−2=0 \begin{aligned} & 10{{x}^{2}}+x\left( -2g-6f-18 \right)+9+6f+c=0 \\\ & \Rightarrow 10{{x}^{2}}+x\left( -2f+6-6f-18 \right)+9+6f+6f-11=0 \\\ & \Rightarrow 10{{x}^{2}}-x\left( 8f+12 \right)+12f-2=0 \\\ \end{aligned}
Now, as the line 3x−y−3=03x-y-3=0 is tangent to the circle so, above the quadratic equation will have only one distinct real root or both roots will be equal so, the discriminant will be zero. Then,
(8f+12)2−4×10(12f−2)=0 ⇒64f2+144+192f−480f+80=0 ⇒64f2−288f+224=0 ⇒2f2−9f+7=0 ⇒2f2−2f−7f+7=0 ⇒2f(f−1)−7(f−1)=0 ⇒(2f−7)(f−1)=0 ⇒f=1,3.5 \begin{aligned} & {{\left( 8f+12 \right)}^{2}}-4\times 10\left( 12f-2 \right)=0 \\\ & \Rightarrow 64{{f}^{2}}+144+192f-480f+80=0 \\\ & \Rightarrow 64{{f}^{2}}-288f+224=0 \\\ & \Rightarrow 2{{f}^{2}}-9f+7=0 \\\ & \Rightarrow 2{{f}^{2}}-2f-7f+7=0 \\\ & \Rightarrow 2f\left( f-1 \right)-7\left( f-1 \right)=0 \\\ & \Rightarrow \left( 2f-7 \right)\left( f-1 \right)=0 \\\ & \Rightarrow f=1,3.5 \\\ \end{aligned}
Now, we got two values of ff so, we will find the value of gg and cc for each value of ff . Then,
f=1 ∵g=f−3 ⇒g=−2 ∵c=6f−11 ⇒c=−5 \begin{aligned} & f=1 \\\ & \because g=f-3 \\\ & \Rightarrow g=-2 \\\ & \because c=6f-11 \\\ & \Rightarrow c=-5 \\\ \end{aligned}
f=3.5 ∵g=f−3 ⇒g=0.5 ∵c=6f−11 ⇒c=10 \begin{aligned} & f=3.5 \\\ & \because g=f-3 \\\ & \Rightarrow g=0.5 \\\ & \because c=6f-11 \\\ & \Rightarrow c=10 \\\ \end{aligned}
Now, there will two such circles and their equation is written below:
g=−2,f=1,c=−5 ⇒x2+y2+4x−2y−5=0 g=0.5,f=3.5,c=10 ⇒x2+y2−x−7y+10=0 \begin{aligned} & g=-2,f=1,c=-5 \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}+4x-2y-5=0 \\\ & g=0.5,f=3.5,c=10 \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}-x-7y+10=0 \\\ \end{aligned}
Thus, the equation of the required circle will be x2+y2+4x−2y−5=0{{x}^{2}}+{{y}^{2}}+4x-2y-5=0 , x2+y2−x−7y+10=0{{x}^{2}}+{{y}^{2}}-x-7y+10=0.

Note: Here, the student should proceed by the basic equation of the circle. After that, try to form the equations correctly as per the given data in the question. Moreover, make some equations before substituting the value of y=3x−3y=3x-3 in the equation of the circle, so that further calculation will be smooth. And students should avoid making calculation mistakes while solving to get the correct answer.