Find the equation of the line inclined at $${{45}^{\circ }}$$to the axis, such that the two circles $${{x}^{2}}+{{y}^{2}}=4,{{x}^{2}}+{{y}^{2}}-10x-14y+65=0$$ intercept equal lengths on it.

Question

Vedantu Content Team · Accepted Answer

Hint:First, before proceeding for this, we must know the equation of the circle where (h, k) is the centre and r is radius as $ {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}} $ . Then, we know that the equation of the line where m is slope and c is y-intercept is given by $ y=mx+c $. Then, it is mentioned in the question that these two chords of two circles are equal which gives the relation as $ {{d}_{1}}={{d}_{2}} $ and by using it, we get the equation of the line which is required.

Complete step by step answer:

In this question, we are supposed to find the equation of the line inclined at \[{{45}^{\circ }}\]to the axis, such that the two circles \[{{x}^{2}}+{{y}^{2}}=4,{{x}^{2}}+{{y}^{2}}-10x-14y+65=0\] intercept equal lengths on it

So, before proceeding for this, we must know the equation of the circle where (h, k) is the centre and r is radius as:

$ {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}} $

Now, by comparing the two equations of the circle given to the standard equation, we could get the centre and radius of both the circles as:

So, for the circle equation \[{{x}^{2}}+{{y}^{2}}=4\], centre is (0, 0) and radius is given by

$ \begin{align}

& {{r}^{2}}=4 \\

& \Rightarrow r=\sqrt{4} \\

& \Rightarrow r=2 \\

\end{align}$

Similarly, for the circle equation \[{{x}^{2}}+{{y}^{2}}-10x-14y+65=0\], we use the completing the square technique and by adding and subtracting 25 and 49, we get:

\[\begin{align}

& {{x}^{2}}+{{y}^{2}}-10x-14y+65+25-25+49-49=0 \\

& \Rightarrow {{x}^{2}}+{{y}^{2}}-10x-14y+65+{{5}^{2}}-{{5}^{2}}+{{7}^{2}}-{{7}^{2}}=0 \\

& \Rightarrow {{\left( x-5 \right)}^{2}}+{{\left( y-7 \right)}^{2}}=-65+25+49 \\

& \Rightarrow {{\left( x-5 \right)}^{2}}+{{\left( y-7 \right)}^{2}}=9 \\

& \Rightarrow {{\left( x-5 \right)}^{2}}+{{\left( y-7 \right)}^{2}}={{3}^{2}} \\

\end{align}\]

So, we get the centre for this circle as (5, 7) and radius as 3.

Now, we will draw the two circles with centres and radius found above and a line passing through them and marked $ {{d}_{1}} $ as base of first circle triangle, $ {{p}_{1}} $ as perpendicular of first circle triangle, $ {{d}_{2}} $ as base of second circle triangle and $ {{p}_{2}} $ as perpendicular of second circle triangle as:

Now, we know that equation of the line where m is slope and c is y-intercept is given by:

$ y=mx+c $

Then, in the question the slope angle is given as $ {{45}^{\circ }} $ , so we get the value of slope as:

$ \begin{align}

& m=\tan {{45}^{\circ }} \\

& \Rightarrow m=1 \\

\end{align} $

Then, by substituting the value of slope as 1, we get the equation of line as:

$ \begin{align}

& y=x+c \\

& \Rightarrow y-x=c \\

\end{align} $

Now, the length of perpendicular $ {{p}_{1}} $ for the equation of above line is given by:

$ \begin{align}

& {{p}_{1}}=\left| \dfrac{-c}{\sqrt{2}} \right| \\

& \Rightarrow {{p}_{1}}=\dfrac{c}{\sqrt{2}} \\

\end{align} $

Now, we can apply Pythagoras theorem in the triangle of first circle and then, we get:

$ \begin{align}

& {{2}^{2}}={{p}_{1}}^{2}+{{d}_{1}}^{2} \\

& \Rightarrow {{d}_{1}}^{2}=4-{{\left( \dfrac{c}{\sqrt{2}} \right)}^{2}} \\

& \Rightarrow {{d}_{1}}=\sqrt{4-{{\left( \dfrac{c}{\sqrt{2}} \right)}^{2}}} \\

\end{align} $

Then, the perpendicular $ {{p}_{2}} $ is given by:

$ \begin{align}

& {{p}_{2}}=\left| \dfrac{7-5-c}{\sqrt{{{1}^{2}}+{{1}^{2}}}} \right| \\

& \Rightarrow {{p}_{2}}=\left( \dfrac{2-c}{\sqrt{2}} \right) \\

\end{align} $

Similarly, can apply Pythagoras theorem in the triangle of second circle and then, we get:

$ \begin{align}

& {{3}^{2}}={{p}_{2}}^{2}+{{d}_{2}}^{2} \\

& \Rightarrow {{d}_{2}}^{2}=9-{{\left( \dfrac{2-c}{\sqrt{2}} \right)}^{2}} \\

& \Rightarrow {{d}_{2}}=\sqrt{9-{{\left( \dfrac{2-c}{\sqrt{2}} \right)}^{2}}} \\

\end{align} $

Now, it is mentioned in the question that these two chords of two circles are equal which gives the relation as:

$ {{d}_{1}}={{d}_{2}} $

So, by substituting their values and also squaring both sides, we get:

\[\begin{align}

& 4-\dfrac{{{c}^{2}}}{2}=9-\dfrac{{{(2-c)}^{2}}}{2} \\

& \Rightarrow 8-{{c}^{2}}=18-\left( 4+{{c}^{2}}-4c \right) \\

& \Rightarrow 8-{{c}^{2}}=18-4-{{c}^{2}}+4c \\

& \Rightarrow -4c=6 \\

& \Rightarrow c=\dfrac{-6}{4} \\

& \Rightarrow c=\dfrac{-3}{2} \\

\end{align}\]

Now, by substituting the value of c in the above found equation of the line as $ y-x=c $ ,we get:

$ \begin{align}

& y-x=\dfrac{-3}{2} \\

& \Rightarrow 2y-2x=-3 \\

& \Rightarrow 2x-2y-3=0 \\

\end{align} $

Hence, we get the required equation of line as $ 2x-2y-3=0 $.

Note:

Now, to solve these type of questions we need to know some of the basics by using the Pythagoras theorem for the right angles triangle which states that the square of the hypotenuse h is equal to the sum of the square of perpendicular p and base b as:

$ {{h}^{2}}={{p}^{2}}+{{b}^{2}} $

Find the equation of the line inclined at \[{{45}^{\circ }}\]to the axis, such that the two circles \[{{x}^{2}}+{{y}^{2}}=4,{{x}^{2}}+{{y}^{2}}-10x-14y+65=0\] intercept equal lengths on it.