The point where the line 4x - 3y + 7 = 0 touc | vedclass

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(C) Let $(a, b)$ be the point where the line $4x - 3y + 7 = 0$ touches the circle $x^2 + y^2 - 6x + 4y - 12 = 0$.The equation of the tangent at $(a, b

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$(-1, 1)$

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(C) Let $(a, b)$ be the point where the line $4x - 3y + 7 = 0$ touches the circle $x^2 + y^2 - 6x + 4y - 12 = 0$.The equation of the tangent at $(a, b)$ to the circle is given by $xa + yb - 3(x + a) + 2(y + b) - 12 = 0$.Rearranging the terms,we get $(a - 3)x + (b + 2)y - (3a - 2b + 12) = 0$.Comparing this with the given line $4x - 3y + 7 = 0$,we have the ratio of coefficients:$\frac{a - 3}{4} = \frac{b + 2}{-3} = \frac{-(3a - 2b + 12)}{7} = k$.From this,$a = 4k + 3$ and $b = -3k - 2$.Substituting these into the third ratio: $-(3(4k + 3) - 2(-3k - 2) + 12) = 7k$.$-(12k + 9 + 6k + 4 + 12) = 7k$ $\Rightarrow -(18k + 25) = 7k$ $\Rightarrow -25k = 25$ $\Rightarrow k = -1$.Substituting $k = -1$ back into the expressions for $a$ and $b$:$a = 4(-1) + 3 = -1$ and $b = -3(-1) - 2 = 1$.Thus,the point of contact is $(-1, 1)$.

The point where the line $4x - 3y + 7 = 0$ touches the circle $x^2 + y^2 - 6x + 4y - 12 = 0$ is

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