Q50 / 160 · AP EAMCET 2020 (18 Sep Shift 2) · Mathematics
The equation of the circle with centre \((2,3)\) and touching the line \(3 x-4 y+1=0\) is
- A\(x^2+y^2+4 x+4 y+12=0\)
- B\(x^2+y^2-4 x-6 y-14=0\)
- C\(x^2+y^2-4 x-6 y+14=0\)
- Correct answerD\(x^2+y^2-4 x-6 y+12=0\)
Solution
Centre \(c=(2,3)\)
radius \(=\) Perpendicular distance from centre \((2,3)\) to the line \(3 x-4 y+1=0\)
\(\begin{aligned} & r=\frac{|3(2)-4(3)+1|}{\sqrt{3^2+(-4)^2}}=\frac{|7-12|}{\sqrt{25}} \\ & r=\frac{5}{5}=1 \end{aligned}\) Equation of circle is \(\begin{aligned} (x-2)^2+(y-3)^2 & =(1)^2 \\ x^2+4-4 x+y^2+9-6 y & =1 \\ x^2+y^2-4 x-6 y+13-1 & =0 \\ x^2+y^2-4 x-6 y+12 & =0 \end{aligned}\) Hence, option (d) is correct.
\(\begin{aligned} & r=\frac{|3(2)-4(3)+1|}{\sqrt{3^2+(-4)^2}}=\frac{|7-12|}{\sqrt{25}} \\ & r=\frac{5}{5}=1 \end{aligned}\) Equation of circle is \(\begin{aligned} (x-2)^2+(y-3)^2 & =(1)^2 \\ x^2+4-4 x+y^2+9-6 y & =1 \\ x^2+y^2-4 x-6 y+13-1 & =0 \\ x^2+y^2-4 x-6 y+12 & =0 \end{aligned}\) Hence, option (d) is correct.
Hence, the correct answer is option: 'D'.