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  • <img alt="\mathrm{\text { If the tangent at }(3,-4) \text { to the circle } x^2+y^2-4 x+2 y-5=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Ctext%20%7B%20If%20the%20tangent

\mathrm{\text { If the tangent at }(3,-4) \text { to the circle } x^2+y^2-4 x+2 y-5=0} \mathrm{\text { cuts the circle } x^2+y^2+16 x+2 y+10=0 \text { in } A \text { and } B \text { then the }}  midpoint of  AB is 

Option: 1

(-6,-7)


Option: 2

(2,-1)


Option: 3

(2,10)


Option: 4

(5,4)


Answers (1)

best_answer

Equation of the tangent to circle \mathrm{x^2+y^2-4 x+2 y-5=0} at point (3, –4) is 3x – 4y – 2(x + 3) + (y – 4) – 5 = 0

⇒ x – 3y – 15 = 0 ...(i)

Centre of the second circle is C(–8, –1).

\mathrm{\text { Slope of line (i) is } \frac{1}{3} \text {. }}

Slope of CM is –3.
Now, equation of CM is y + 1 = –3(x + 8)
⇒ 3x + y + 25 = 0 ...(ii)
Solving (i) and (ii), we get M ≡(–6, –7)

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SANGALDEEP SINGH

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