If m is the slope and P(8, beta) is the midpo | vedclass

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(C) The equation of the circle is $x^2+y^2=125$. Given $P(8, \beta)$ is the midpoint of the chord. The equation of the chord with midpoint $(x_1, y_1)

Vedclass · Accepted Answer

$6$

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(C) The equation of the circle is $x^2+y^2=125$. Given $P(8, \beta)$ is the midpoint of the chord. The equation of the chord with midpoint $(x_1, y_1)$ is $T=S_1$,which is $xx_1+yy_1=x_1^2+y_1^2$. Substituting $(8, \beta)$,we get $8x+\beta y = 64+\beta^2$,or $8x+\beta y - (64+\beta^2) = 0$. The slope of this chord is $m = -\frac{8}{\beta}$. For $m$ to be an integer,$\beta$ must be a divisor of $8$. Thus,$\beta \in \{ \pm 1, \pm 2, \pm 4, \pm 8 \}$. Since the point $P(8, \beta)$ must lie inside the circle,$8^2+\beta^2 Checking the values: If $\beta = \pm 1$,$\beta^2 = 1 If $\beta = \pm 2$,$\beta^2 = 4 If $\beta = \pm 4$,$\beta^2 = 16 If $\beta = \pm 8$,$\beta^2 = 64 > 61$ (Invalid). Thus,the possible values for $\beta$ are $\pm 1, \pm 2, \pm 4$,which gives a total of $6$ values.

If $m$ is the slope and $P(8, \beta)$ is the midpoint of a chord of contact of the circle $x^2+y^2=125$,then the number of values of $\beta$ such that $\beta$ and $m$ are integers is

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