A circle passes through the point $\left( 3,\sqrt{\dfrac{7}{2}} \right)$ and touches the line pair ${{x}^{2}}-{{y}^{2}}-2x+1=0$ . The coordinates of the centre of circle are:
a) (4, 0)
b) (5, 0)
c) (6, 0)
d) (0, 4)
VerifiedHint: Find the equations of tangents by factoring the given equation of pair of straight lines using the identity,
${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
Use the concept of tangent and radius of any circle that the perpendicular length of any tangent from the centre of the circle is equal to the radius of the circle. Standard equation of circle with centre
$\left( {{x}_{1}},{{y}_{1}} \right)$ and radius ‘r’ is
${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}$
Complete step-by-step answer:
Let us suppose the equation of the circle as
${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{R}^{2}}..............\left( i \right)$
Which represents the standard equation of circle with centre $\left( {{x}_{1}},{{y}_{1}} \right)$ and radius of ‘R’. As it is given in the question that circle passes through the point $\left( 3,\sqrt{\dfrac{7}{2}} \right)$ and touching the line pair ${{x}^{2}}-{{y}^{2}}-2x+1=0$ .
Here, ${{x}^{2}}-{{y}^{2}}-2x+1=0$ is consisting of two line so, we need to factorize the relation given in form of pair of straight lines to get the equations of both the lines individually. So, we have
$ {{x}^{2}}-{{y}^{2}}-2x+1=0 $
$ \Rightarrow \left( {{x}^{2}}-2x+1 \right)-{{y}^{2}}=0 $
$ \left( {{\left( x \right)}^{2}}-2\times 1\times x+{{1}^{2}} \right)-{{y}^{2}}=0................\left( ii \right) $
Now, we know the algebraic identity
${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}.................\left( iii \right)$
So, we can compare the RHS of equation (iii) with ${{\left( x \right)}^{2}}-2\times 1\times x+{{\left( 1 \right)}^{2}}$ and hence we get that this expression can be replace by ${{\left( x-1 \right)}^{2}}$ . So, we get equation (ii) as
$ {{\left( x-1 \right)}^{2}}-{{y}^{2}}=0 $
$ \Rightarrow {{\left( x-1 \right)}^{2}}-{{\left( y \right)}^{2}}=0.....................\left( iv \right) $
Now, relate this expression with the algebraic identity ${{a}^{2}}-{{b}^{2}}$ , which is given as
${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right).............\left( v \right)$
So, we can get equation (iv) as
$ {{\left( x-1 \right)}^{2}}-{{y}^{2}}=0 $
$ \left( x-1-y \right)-\left( x-1+y \right)=0 $
Hence, we get equation of both lines as
x – 1 – y = 0 and x – 1 + y = 0
x – y = 1 and x + y = 1
As, these two lines are touching the circle given in equation (i). So, we can draw diagram as
Now, we have equation of two tangents i.e. AP and BP and a point $\left( 3,\sqrt{\dfrac{7}{2}} \right)$ on the circle.
As we can notice that the OA and OB are the perpendicular lengths from point O to the tangents AP and BP, because tangent at any point is perpendicular to the radius to that point. Now, we know the perpendicular length of any line Ax + By + c = 0 from any point $\left( {{x}_{1}},{{y}_{1}} \right)$ can be given by the relation
$\text{perpendicular distance =}\left| \dfrac{A{{x}_{1}}+B{{y}_{1}}+c}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right|..............\left( vi \right)$
So, lengths OA = R and OB = R from the point $O\left( {{x}_{1}},{{y}_{1}} \right)$ to the tangents AP (x – y – 1 = 0) and BP (x + y – 1 = 0). So, we can get equations as
$ OA=R=\left| \dfrac{{{x}_{1}}-{{y}_{1}}-1}{\sqrt{{{1}^{2}}+{{1}^{2}}}} \right|=\left| \dfrac{{{x}_{1}}-{{y}_{1}}-1}{\sqrt{2}} \right|..................\left( vii \right) $
$ OB=R=\left| \dfrac{{{x}_{1}}+{{y}_{1}}-1}{\sqrt{{{1}^{2}}+{{1}^{2}}}} \right|=\left| \dfrac{{{x}_{1}}+{{y}_{1}}-1}{\sqrt{2}} \right|..................\left( viii \right) $
So, we can equate both the equations i.e (vii) and (viii). So, we get equation as
$\left| \dfrac{{{x}_{1}}-{{y}_{1}}-1}{\sqrt{2}} \right|=\left| \dfrac{{{x}_{1}}+{{y}_{1}}-1}{\sqrt{2}} \right|$
Squaring on both sides, we get
$ \dfrac{{{\left( {{x}_{1}}-{{y}_{1}}-1 \right)}^{2}}}{{{\left( \sqrt{2} \right)}^{2}}}=\dfrac{{{\left( {{x}_{1}}+{{y}_{1}}-1 \right)}^{2}}}{{{\left( \sqrt{2} \right)}^{2}}} $
$ {{\left( {{x}_{1}}-{{y}_{1}}-1 \right)}^{2}}={{\left( {{x}_{1}}+{{y}_{1}}-1 \right)}^{2}} $
Now, use the identity given below
${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}-2ab-2bc-2ca$
So, we get
$ x_{1}^{2}+y_{1}^{2}+1-2{{x}_{1}}{{y}_{1}}+2{{y}_{1}}-2{{x}_{1}} $
$ =x_{1}^{2}+y_{1}^{2}+1+2{{x}_{1}}{{y}_{1}}-2{{y}_{1}}-2{{x}_{1}} $
$ \Rightarrow 4{{x}_{1}}-4{{y}_{1}}=0 $
$ 4{{y}_{1}}\left( {{x}_{1}}-1 \right)=0 $
So, we get
$ {{y}_{1}}=0 $
$ \Rightarrow {{x}_{1}}=1 $
Now, as the distance OC is represented as radius also, so we can get radius by applying distance formula to it. So, we have distance formula as
$D=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$
Where D is the distance between points
$\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$
So, we get
$OC=R=\sqrt{{{\left( {{x}_{1}}-3 \right)}^{2}}+{{\left( {{y}_{1}}-\sqrt{\dfrac{7}{2}} \right)}^{2}}}.......................\left( ix \right)$
Now, equate the above equation with (viii) as both are representing the radius of circle
$R=\sqrt{{{\left( {{x}_{1}}-3 \right)}^{2}}+{{\left( {{y}_{1}}-\sqrt{\dfrac{7}{2}} \right)}^{2}}}=\left| \dfrac{{{x}_{1}}-{{y}_{1}}-1}{\sqrt{2}} \right|.....................\left( \text{x} \right)$
Put ${{y}_{1}}=0$ to the above expression, we get
$ \sqrt{{{\left( {{x}_{1}}-3 \right)}^{2}}+{{\left( 0-\sqrt{\dfrac{7}{2}} \right)}^{2}}}=\left| \left( \dfrac{{{x}_{1}}-0-1}{\sqrt{2}} \right) \right| $
$ \sqrt{{{\left( {{x}_{1}}-3 \right)}^{2}}+\dfrac{7}{2}}=\left| \dfrac{{{x}_{1}}-1}{\sqrt{2}} \right| $
Squaring on both the sides, we get
$ {{\left( {{x}_{1}}-3 \right)}^{2}}+\dfrac{7}{2}=\dfrac{{{\left( {{x}_{1}}-1 \right)}^{2}}}{2} $
$ x_{2}^{2}+9-6{{x}_{1}}+\dfrac{7}{2}=\dfrac{x_{1}^{2}+1-2{{x}_{1}}}{2} $
$ \Rightarrow \dfrac{2x_{1}^{2}+18-12{{x}_{1}}+7}{2}=\dfrac{x_{1}^{2}+1-2{{x}_{1}}}{2} $
$ \Rightarrow 2x_{1}^{2}-12{{x}_{1}}+25=x_{1}^{2}-2{{x}_{1}}+1 $
$ x_{1}^{{}}-10{{x}_{1}}+24=0 $
Now, on factoring the above equation, we get
$ x_{1}^{2}-6{{x}_{1}}-4{{x}_{1}}+24=0 $
$ {{x}_{1}}\left( {{x}_{1}}-6 \right)-4\left( {{x}_{1}}-6 \right)=0 $
$ \left( {{x}_{1}}-4 \right)\left( {{x}_{1}}-6 \right)=0 $
So, we get
$ {{x}_{1}}=4 $
$ \Rightarrow {{x}_{1}}=6 $
Hence, we get a possible centre of the circle as (6, 0) or (4, 0).
Similarly, put ${{x}_{1}}=1$ to the equation (x). So, we get
$ \sqrt{{{\left( 1-3 \right)}^{2}}+{{\left( {{y}_{1}}-\sqrt{\dfrac{7}{2}} \right)}^{2}}}=\left| \dfrac{1-{{y}_{1}}-1}{\sqrt{2}} \right| $
$ \sqrt{4+{{\left( {{y}_{1}}-\sqrt{\dfrac{7}{2}} \right)}^{2}}}=\left| \dfrac{-{{y}_{1}}}{\sqrt{2}} \right| $
Squaring on both the sides, we get
$ 4+{{\left( {{y}_{1}}-\sqrt{\dfrac{7}{2}} \right)}^{2}}=\dfrac{y_{1}^{2}}{2} $
$ 4+y_{1}^{2}+\dfrac{7}{2}-2{{y}_{1}}\sqrt{\dfrac{7}{2}}=\dfrac{y_{1}^{2}}{2} $
$ \dfrac{y_{1}^{2}}{2}-2{{y}_{1}}\sqrt{\dfrac{7}{2}}+\dfrac{15}{2}=0 $
Now, discriminant $\left( ={{b}^{2}}-4ac \right)$ of the above equation can be given as
$D={{\left( -2\sqrt{\dfrac{7}{2}} \right)}^{2}}-4\times \dfrac{1}{2}\times \dfrac{15}{2}$
D = 14 – 15 = -1
D < 0 So, roots of this equation will not be real. Hence, ${{x}_{1}}=1$ is not possible. So, we have possible value of centre are (6, 0) and (4, 0)
So, the correct answers are “Option A and C”.
Note: Don’t use the perpendicular distance without mod, as one may not get some values of $\left( {{x}_{1}},{{y}_{1}} \right)$ or centre in the solution. The relation,
$\left| \dfrac{{{x}_{1}}+{{y}_{1}}-1}{\sqrt{2}} \right|=\left| \dfrac{{{x}_{1}}+{{y}_{1}}-1}{\sqrt{2}} \right|$
Will give only value i.e. ${{x}_{1}}=0$ , if one may miss the mod sign to the expression (perpendicular distance). So, take care of it with these kinds of questions.
One may get confused with two values of the centre of the circle. Two values of centre suggests that there will be two possible circles with given information in the problem.
