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Find the equations of the tangents to the circle `x^2+y^2-6x+4y=12`which are parallel to the straight line`4x+3y+5=0` |
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Answer» Given circle is `x^(2)+y^(2)-6x+4y-12=0` Centre of the circle is `(3,-2)` and radius is `sqrt((-3)^(2)+(2)^(2)-(-12))=5` Given that tangents are parallel to the line `4x+3y+5=0`. Therefore, the equations of tangents take the form `4x+3y+c=0`. Distance of centre of the circle from the tangent line is radius. `:. |(4(3)+3(-2)+k)/(sqrt(16+9))|=5` `implies 6+k=+-25` `implies k=19` and `-31` Therefore, the equations of required tangents are `4x+3y+19=0` and `4x+3y-31=0` Alternative method `:` Given circle in center-radius form form is `(x-3)^(2)+(y+2)^(2)=5^(2)` Tangents are parallel to the line `4x+3y+5=0` . `:. ` Slope of tangents `=-(4)/(3)` Hence, using the equation of tangents `y+f=m(x+g)+-sqrt(m^(2)+1)sqrt(g^(2)+f^(2)-c)`, required equations of tangents are obtained as `y+2=-(4)/(3)(x-3)+-5sqrt((-(4)/(3))^(2)+1)` or `4x+3y+19=0` and `4x+3y-31=0` |
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