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BITSAT 2022 Paper

Section: Mathematics

Question : 118 of 130

Marks: +1, -0

4x-5y=20

x^2+y^2=9

Solution: 👈: Video Solution

Let

P\left(t, 4t-\frac{20}{5}\right)

be a point on the line

4x-5y=20

. Then the chord of contact of tangents drawn from

P

to the circle

x^2+y^2=9

tx\left(4t-\frac{20}{5}\right)y=9 \ldots (i)

Let

\theta(h, k)

be the mid-point of this chord of contact, then its equation is also

hx+ky=h^2+k^2 \ldots (ii)

Clearly Eqs. (i) and (ii) represent the same line

\therefore \frac{t}{h} = \frac{4t-20}{5k} = \frac{9}{h^2+k^2}

\Rightarrow \frac{t}{h} = \frac{9}{h^2+k^2} \text{ and } \frac{t}{h} = \frac{4t-20}{5k}

\Rightarrow t = \frac{9h}{h^2+k^2} \text{ and } t = \frac{20h}{4h-5k}

\Rightarrow \frac{9h}{h^2+k^2} = \frac{20h}{4h-5k}

\Rightarrow h\left(20(h^2+k^2)-36h+45k\right)=0

x=0, \text{ or } \left[20(x^2+y^2)-36x+45y\right]=0

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