Question

If four distinct points $\left( 2,3 \right)$, $\left( 0,2 \right)$, $\left( 4,5 \right)$ and $\left( 0,t \right)$ are concyclic, then ${{t}^{3}}+17$ is equal to

Answer 1

Hint: First of all, we will substitute the values of points $\left( 2,3 \right)$, $\left( 0,2 \right)$, $\left( 4,5 \right)$ in equation of circle i.e. ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. From this we will get three expressions in terms of g, f and c and on solving the expressions we will get values of g, f and c. Again, on substituting the values of g, f and c in the equation of circle we will get a new equation and the point $\left( 0,t \right)$, must satisfy the expression so, on substituting the value of point we will get the value of t. Then from that value of t we will find the value of equation ${{t}^{3}}+17$.

Complete step-by-step answer:
In question we are told that four distinct points are concyclic and we are asked to find the value equation ${{t}^{3}}+17$, so, first of all we will understand what are concyclic points.
In geometry, a set of points are said to be concyclic (or cocyclic) if they lie on a common circle and all concyclic points are the same distance from the center of the circle. The figure can be seen as,

Now, as the points $\left( 2,3 \right)$, $\left( 0,2 \right)$, $\left( 4,5 \right)$ and $\left( 0,t \right)$ they must satisfy the equation of circle i.e. ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ . Now, we will substitute the values of points $\left( 2,3 \right)$, $\left( 0,2 \right)$, $\left( 4,5 \right)$ in the equation of the circle.
For point $\left( 2,3 \right)$, value of $x=2$ and $y=3$, so on substituting this the equation we will become,
${{2}^{2}}+{{3}^{2}}+2g\left( 2 \right)+2f\left( 3 \right)+c=0$
$\Rightarrow 4+9+4g+6f+c=0$
$\Rightarrow 4g+6f+c=-13$ ……………(i)
For point $\left( 0,2 \right)$, value of $x=0$ and $y=2$, so on substituting this the equation we will become,
${{0}^{2}}+{{2}^{2}}+2g\left( 0 \right)+2f\left( 2 \right)+c=0$
$\Rightarrow 0+4+0+4f+c=0$
$\Rightarrow 4f+c=-4$ …………………..(ii)
For point $\left( 4,5 \right)$, value of $x=2$ and $y=3$, so on substituting this the equation we will become,
${{4}^{2}}+{{5}^{2}}+2g\left( 4 \right)+2f\left( 5 \right)+c=0$
$\Rightarrow 16+25+8g+10f+c=0$
$\Rightarrow 8g+10f+c=-41$ ………………..(iii)
Now, on evaluating the expression (ii) we will get,
$4f+c=-4$
$\Rightarrow c=-4-4f$ ………………..(iv)
Now, on substituting the value of expression (iv) in expression (i) we will get,
$4g+6f-4-4f=-13$
$\Rightarrow 4g+6f-4f=-13+4$
$\Rightarrow 4g+2f=-9$ ……………..(v)
Now, on substituting the value of (iv) in expression (iii) we will get,
$8g+10f-4-4f=-41$
$\Rightarrow 8g+10f-4f=-41+4$
$\Rightarrow 8g+6f=-37$ ……………..(vi)
Now, on multiplying the expression (v) with 2 and simplifying further we will get,
$\Rightarrow 8g=-18-4f$ ……………..(vii)
Now, on substituting the value of (vii) in expression (vi) we will get,
$-18-4f+6f=-37$
$\Rightarrow 2f=-37+18\Rightarrow f=\dfrac{-19}{2}$
Now, substituting this value of f in expression (vii) we will get,
$8g=-18-4\left( \dfrac{-19}{2} \right)$
$\Rightarrow 8g=-18-2\left( -19 \right)\Rightarrow g=\dfrac{-18+38}{8}$
$\Rightarrow g=\left( \dfrac{20}{8} \right)=\dfrac{5}{2}$
Now, substituting the value of f in expression (iv) we will get,
$c=-4-4\left( \dfrac{-19}{2} \right)$
$\Rightarrow c=-4-2\left( -19 \right)\Rightarrow c=-4+38=34$
So, the values are $g=\dfrac{5}{2}$, $f=-\dfrac{19}{2}$ and $c=34$.
Now, on substituting these values the in equation of circle we will get,
${{x}^{2}}+{{y}^{2}}+2\left( \dfrac{5}{2} \right)x+2\left( \dfrac{-19}{2} \right)y+34=0$ …………………(viii)
Now, as the circle passes through the point $\left( 0,t \right)$, on substituting the value in expression (viii) we will get,
${{0}^{2}}+{{t}^{2}}+2\left( \dfrac{5}{2} \right)0+2\left( \dfrac{-19}{2} \right)t+34=0$
$\Rightarrow {{t}^{2}}-19t+34=0$
Now, we know that 34 can be written as, $34=2\times 17$, so, we will write $19t$ as $17t+2t$, so on substituting this value we will get,
$\Rightarrow {{t}^{2}}-\left( 17t+2t \right)+34=0$
$\Rightarrow {{t}^{2}}-17t-2t+34=0$
On, simplifying further we will get,
$\Rightarrow t\left( t-17 \right)-2\left( t-17 \right)=0$
$\Rightarrow \left( t-2 \right)\left( t-17 \right)=0$
$\left( t-2 \right)=0\Rightarrow t=2$ and $\left( t-17 \right)=0\Rightarrow t=17$
But, if we consider the value of t as 2 then the point will be $\left( 0,2 \right)$, which already exists so the value of t will be 17, which gives, $\left( 0,17 \right)$.
Now, substituting the value of t in expression ${{t}^{3}}+17$, we will get,
${{t}^{3}}+17={{17}^{3}}+17=4930$
Hence, the value of ${{t}^{3}}+17$ is 4930.

Note: The equation of circle is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$, but instead of this if student consider the equation as ${{x}^{2}}+{{y}^{2}}+gx+fy+c=0$ and substitute the value accordingly then we will get,
${{2}^{2}}+{{3}^{2}}+2g+3f+c=0$ for point $\left( 2,3 \right)$, which is different from our expression i.e. ${{2}^{2}}+{{3}^{2}}+2g\left( 2 \right)+2f\left( 3 \right)+c=0$. So, by considering the wrong equation the whole expression changes and due to that values will also change and the answer will be wrong at the end. So, students must remember the equation of the circle and use it properly to avoid mistakes.