Question:
Suppose the tangents drawn to the circle \(x^2+y^2-6x-4y-11=0\) from \(P(1,8)\) touch the circle at \(A\) and \(B\). Then the centre of the circle passing through \(P,A\) and \(B\) is
Suppose the tangents drawn to the circle \(x^2+y^2-6x-4y-11=0\) from \(P(1,8)\) touch the circle at \(A\) and \(B\). Then the centre of the circle passing through \(P,A\) and \(B\) is
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For tangents drawn from an external point to a circle, the chord joining the points of contact is found using \(T=0\).
Updated On: Jun 25, 2026
- \((2,5)\)
- \((-2,-5)\)
- \((-2,5)\)
- \((2,-5)\)
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The Correct Option is A
Solution and Explanation
Step 1: Write the given circle.
Given circle is \[ S\equiv x^2+y^2-6x-4y-11=0 \] Here, \[ g=-3,\quad f=-2,\quad c=-11 \]
Step 2: Find the chord of contact from \(P(1,8)\).
The chord of contact from \((x_1,y_1)\) to the circle \[ x^2+y^2+2gx+2fy+c=0 \] is given by \[ T=0 \] So, \[ xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0 \] Substituting \(x_1=1,\ y_1=8,\ g=-3,\ f=-2,\ c=-11\): \[ x+8y-3(x+1)-2(y+8)-11=0 \] \[ x+8y-3x-3-2y-16-11=0 \] \[ -2x+6y-30=0 \] Dividing by \(-2\), \[ x-3y+15=0 \] Thus, line \(AB\) is \[ L\equiv x-3y+15=0 \]
Step 3: Write the family of circles passing through \(A\) and \(B\).
The circle passing through the intersection points of \[ S=0 \] and \[ L=0 \] is \[ S+\lambda L=0 \] So, \[ x^2+y^2-6x-4y-11+\lambda(x-3y+15)=0 \]
Step 4: Use the condition that the circle passes through \(P(1,8)\).
Substitute \((1,8)\) in \(S\): \[ S(1,8)=1+64-6-32-11 \] \[ S(1,8)=16 \] Now substitute \((1,8)\) in \(L\): \[ L(1,8)=1-24+15 \] \[ L(1,8)=-8 \] Since \(P\) lies on the required circle: \[ S(P)+\lambda L(P)=0 \] \[ 16+\lambda(-8)=0 \] \[ \lambda=2 \]
Step 5: Find the required circle.
Substitute \(\lambda=2\): \[ x^2+y^2-6x-4y-11+2(x-3y+15)=0 \] \[ x^2+y^2-6x-4y-11+2x-6y+30=0 \] \[ x^2+y^2-4x-10y+19=0 \]
Step 6: Find the centre.
For \[ x^2+y^2+2gx+2fy+c=0, \] centre is \[ (-g,-f) \] Here, \[ 2g=-4 \Rightarrow g=-2 \] and \[ 2f=-10 \Rightarrow f=-5 \] So, centre is \[ (2,5) \]
Step 7: Final conclusion.
Therefore, the centre of the required circle is \[ \boxed{(2,5)} \]
Given circle is \[ S\equiv x^2+y^2-6x-4y-11=0 \] Here, \[ g=-3,\quad f=-2,\quad c=-11 \]
Step 2: Find the chord of contact from \(P(1,8)\).
The chord of contact from \((x_1,y_1)\) to the circle \[ x^2+y^2+2gx+2fy+c=0 \] is given by \[ T=0 \] So, \[ xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0 \] Substituting \(x_1=1,\ y_1=8,\ g=-3,\ f=-2,\ c=-11\): \[ x+8y-3(x+1)-2(y+8)-11=0 \] \[ x+8y-3x-3-2y-16-11=0 \] \[ -2x+6y-30=0 \] Dividing by \(-2\), \[ x-3y+15=0 \] Thus, line \(AB\) is \[ L\equiv x-3y+15=0 \]
Step 3: Write the family of circles passing through \(A\) and \(B\).
The circle passing through the intersection points of \[ S=0 \] and \[ L=0 \] is \[ S+\lambda L=0 \] So, \[ x^2+y^2-6x-4y-11+\lambda(x-3y+15)=0 \]
Step 4: Use the condition that the circle passes through \(P(1,8)\).
Substitute \((1,8)\) in \(S\): \[ S(1,8)=1+64-6-32-11 \] \[ S(1,8)=16 \] Now substitute \((1,8)\) in \(L\): \[ L(1,8)=1-24+15 \] \[ L(1,8)=-8 \] Since \(P\) lies on the required circle: \[ S(P)+\lambda L(P)=0 \] \[ 16+\lambda(-8)=0 \] \[ \lambda=2 \]
Step 5: Find the required circle.
Substitute \(\lambda=2\): \[ x^2+y^2-6x-4y-11+2(x-3y+15)=0 \] \[ x^2+y^2-6x-4y-11+2x-6y+30=0 \] \[ x^2+y^2-4x-10y+19=0 \]
Step 6: Find the centre.
For \[ x^2+y^2+2gx+2fy+c=0, \] centre is \[ (-g,-f) \] Here, \[ 2g=-4 \Rightarrow g=-2 \] and \[ 2f=-10 \Rightarrow f=-5 \] So, centre is \[ (2,5) \]
Step 7: Final conclusion.
Therefore, the centre of the required circle is \[ \boxed{(2,5)} \]
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