The incentre of an equilateral triangle is $(1,1)$ and the equation of one side is $3 x+4 y+3=0$ Then, the equation of the circumcircle of the triangle is
- $x^{2}+y^{2}-2x-2y-2=0$
- $x^{2}+y^{2}-2x-2y-14=0$
- $x^{2}+y^{2}-2x-2y+2=0$
- $x^{2}+y^{2}-2x-2y+14=0$
The Correct Option is B
Solution and Explanation
$\therefore G D=$ Length of perpendicular from the point
$G(1,1)$ to the line $3 x+4 y+3=0$
$3 x+4 y+3=0$
$=\frac{3(1)+4(1)+3}{\sqrt{3^{2}+4^{2}}}=2$
$A G=2 G D=4$
$\therefore$ Equation of circumcircle with centre at $(1,1)$ and radius $=4$ units
$(x-1)^{2}+(y-1)^{2} =4^{2}$
$\Rightarrow x^{2}+ y^{2}-2 x-2 y-14 =0$
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Concepts Used:
Equation of a Line in Space
In a plane, the equation of a line is given by the popular equation y = m x + C. Let's look at how the equation of a line is written in vector form and Cartesian form.
Vector Equation
Consider a line that passes through a given point, say ‘A’, and the line is parallel to a given vector '\(\vec{b}\)‘. Here, the line ’l' is given to pass through ‘A’, whose position vector is given by '\(\vec{a}\)‘. Now, consider another arbitrary point ’P' on the given line, where the position vector of 'P' is given by '\(\vec{r}\)'.
\(\vec{AP}\)=𝜆\(\vec{b}\)
Also, we can write vector AP in the following manner:
\(\vec{AP}\)=\(\vec{OP}\)–\(\vec{OA}\)
𝜆\(\vec{b}\) =\(\vec{r}\)–\(\vec{a}\)
\(\vec{a}\)=\(\vec{a}\)+𝜆\(\vec{b}\)
\(\vec{b}\)=𝑏1\(\hat{i}\)+𝑏2\(\hat{j}\) +𝑏3\(\hat{k}\)