Question

Three vertices of a triangle are (3,0), (0,0) and (0,5). Find out the equation of the circle passing through these vertices.
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Answer 1

Question :-

• Three vertices of a triangle are (3,0), (0,0) and (0,5). Find out the equation of the circle passing through these vertices.

$\large \orange{AηsωeR} ✍$

Given :-

Three vertices of a triangle are (3,0), (0,0) and (0,5).

To Find :-

The equation of the circle passing through these vertices.

$\begin{gathered}\Large{\bold{\pink{\underline{CaLcUlAtIoN\::}}}}\end{gathered}$

$\begin{gathered}\bf\red{Let,}\end{gathered}$

The vertices of a triangle be

$\bf \: ⟼ A(3,0)$

$\bf \: ⟼ B(0,0)$

$\bf \: ⟼ C(0,5)$

$\bf \: ⟼ AB \: = 3 \: units$

$\bf \: ⟼ BC = 5 \: units$

Now, we have to find the equation of circle passes through A, B & C.

Construction:-

JoinAC.

$\bf \: ⟼ As \: ∠ABC = 90°.$

$\bf \: ⟼ AC \: is \: diameter \: of \: circle.$

So, Using pythagoras theorem

$\begin{gathered}{\boxed{\bf{\pink{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}}}}\end{gathered}$

$\bf \: ⟼ {BC}^{2} + {AB}^{2} = {AC}^{2}$

$\bf \: ⟼ {AC}^{2} = {3}^{2} + {5}^{2}$

$\bf \: ⟼ {AC}^{2} = 25 + 9$

$\bf \: ⟼ {AC}^{2} = 34$

$\bf \: ⟼ AC = \sqrt{34} \: units$

$\begin{gathered}\bf\red{Let,} \end{gathered}$

$\begin{gathered}\longmapsto\:\:\bf{Radius\:is \:r\:units}. \end{gathered}$

$\bf \: ⟼ r = \dfrac{1}{2} \times AC$

$\bf \: ⟼ r = \dfrac{ \sqrt{34} }{2} \: units$

$\bf \: ⟼ Let \: centre \: of \: circle \: be \: O(h, k).$

Centre O lies on AC.

So, O is the midpoint of AC.

Using Midpoint formula,

$\bf \:( x, y) = (\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} )</p><p>$

$\bf \: ⟼ ( h, k) = (\dfrac{0+3}{2} , \dfrac{5+0}{2} )$

$\bf \: ⟼ ( h, k) = (\dfrac{3}{2} , \dfrac{5}{2} )$

$\begin{gathered}\bf\red{So, \: equation \: of \: circle \: is \: } \end{gathered}$

$\bf \: ⟼ {(x - h)}^{2} + {(y - k)}^{2} = {r}^{2}$

On substituting the values of h, k and r, we get

$\bf \: ⟼ {(x - \dfrac{3}{2}) }^{2} + {(y - \dfrac{5}{2}) }^{2} = {(\dfrac{ \sqrt{34} }{2} )}^{2}$

$\bf \: ⟼ {x}^{2} + \dfrac{9}{4} - 3x + {y}^{2} + \dfrac{25}{4} - 5y = \dfrac{34}{4}$

$\bf \: ⟼ {x}^{2} + {y}^{2} - 3x - 5y = 0$