119709
If \(x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0\) is the smallest circle through the points of intersection of \(x^2+y^2=a^2\) and \(x \cos \alpha+y \sin \alpha=\) \(\mathbf{p}, \mathbf{0}\lt \mathbf{p}\lt \mathbf{a}\), then \(\lambda=\)
1 1
2 \(-\mathrm{p}\)
3 \(-2 \mathrm{p}\)
4 \(-3 \mathrm{p}\)
Explanation:
C Equation of circle \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{a}^2+\lambda(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha-\mathrm{p})=0\) is the smallest circle, then centre \(\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)\) lies on the line \(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{p}\) \(\therefore\) Put centre in the line (i), then we get- \(\frac{-\lambda \cos ^2 \alpha}{2}-\frac{\lambda \sin ^2 \alpha}{2}=\mathrm{p}\) \(\Rightarrow \frac{-\lambda}{2}=\mathrm{p}\) \(\Rightarrow \lambda=-2 \mathrm{p}\)
TS EAMCET-11.09.2020
Conic Section
119710
If the poles of the line \(x-y=0\) with respect the circles \(x^2+y^2-2 g_i x+c_i^2=0(i=1,2,3)\) are \(\left(\alpha_i\right.\), \(\left.\beta_i\right)\), then \(\sum_{i=1}^3 \frac{\alpha_i+\beta_i}{g_i}=\)
1 3
2 6
3 \(\frac{3}{2}\)
4 \(\frac{3}{4}\)
Explanation:
A Since equation polar of point \(\left(\alpha_i, \beta_i\right)\) with respect to the circle \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{~g}_{\mathrm{i}} \mathrm{x}+\mathrm{c}_{\mathrm{i}}^2=0\) is \(\alpha_i x+\beta_i y-g_i\left(x+\alpha_i\right)+c_i^2=0\) \(\left(\alpha_i-g_i\right) x+\beta_i y+c_i^2-\alpha_i g_i=0\) Now, on comparing with line \(x-y=0\), we get \(\frac{\alpha_{\mathrm{i}}-\mathrm{g}_{\mathrm{i}}}{1}=\frac{\beta_{\mathrm{i}}}{-1}=\frac{\mathrm{c}_{\mathrm{i}}^2-\alpha_{\mathrm{i}} \mathrm{g}_{\mathrm{i}}}{0}\) \(\mathrm{c}_{\mathrm{i}}^2=\alpha_{\mathrm{i}} \mathrm{g}_{\mathrm{i}} \text { and } \alpha_{\mathrm{i}}+\beta_{\mathrm{i}}=\mathrm{g}_{\mathrm{i}}\) \(\therefore \quad \sum_{\mathrm{i}=1}^3 \frac{\alpha_{\mathrm{i}}+\beta_{\mathrm{i}}}{\mathrm{g}_{\mathrm{i}}}=\sum_{\mathrm{i}=1}^3(1)=3\)
TS EAMCET-11.09.2020
Conic Section
119711
If the circle \(x^2+y^2=a^2\) intersects the hyperbola \(x y=b^2\) at four points \(\left(x_1, y_1\right),\left(x_2, y_2\right),\left(x_3, y_3\right)\), \(\left(\mathbf{x}_4, \mathbf{y}_4\right)\), then \(\mathbf{y}_1 \mathbf{y}_2 \mathbf{y}_3 \mathbf{y}_4=\)
1 a
2 0
3 \(b^4\)
4 \(b^2\)
Explanation:
C We have, And \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2\) \(\mathrm{xy}=\mathrm{b}^2\) From Eq. (ii) \(x=\frac{b^2}{y}\) From Eq. (i) \(\left(\frac{\mathrm{b}^2}{\mathrm{y}}\right)^2+\mathrm{y}^2=\mathrm{a}^2\) \(b^4+y^4=a^2 y^2\) \(y^4-a^2 y^2+b^4=0\)This is an equation of 4th degree in \(y\) and its four roots are \(\mathrm{y}_1, \mathrm{y}_2, \mathrm{y}_3, \mathrm{y}_4\) then \(\mathrm{y}_1 \mathrm{y}_2 \mathrm{y}_3 \mathrm{y}_4=\mathrm{b}^4\)
TS EAMCET-10.09.2020
Conic Section
119712
If \(\alpha \neq-4\) and \((2, \alpha)\) is the mid-point of a chord of the circle \(x^2+y^2-4 x+8 y+6=0\), then the values of the \(y\)-intercept of the chord lie in the interval
1 \((-4-\sqrt{14},-4+\sqrt{14})\)
2 \((-4,4)\)
3 \((4, \sqrt{14}, 4+\sqrt{14})\)
4 \((-2,2)\)
Explanation:
A We have, \((2, \alpha)\) is mid-point of chord of circle \(x^2+y^2-4 x+8 y+6=0\) \(\because(2, \alpha)\) inside the circle \(\therefore 4+\alpha^2-8+8 \alpha+6 \leq 0\) \(\alpha^2+8 \alpha+2 \leq 0\) \(\alpha=\frac{-8 \pm \sqrt{64-8}}{2}\) \(\alpha=\frac{-8 \pm 2 \sqrt{14}}{2}=-4 \pm \sqrt{14}\) \(\alpha \in(-4-\sqrt{14},-4+\sqrt{14})\)
119709
If \(x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0\) is the smallest circle through the points of intersection of \(x^2+y^2=a^2\) and \(x \cos \alpha+y \sin \alpha=\) \(\mathbf{p}, \mathbf{0}\lt \mathbf{p}\lt \mathbf{a}\), then \(\lambda=\)
1 1
2 \(-\mathrm{p}\)
3 \(-2 \mathrm{p}\)
4 \(-3 \mathrm{p}\)
Explanation:
C Equation of circle \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{a}^2+\lambda(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha-\mathrm{p})=0\) is the smallest circle, then centre \(\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)\) lies on the line \(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{p}\) \(\therefore\) Put centre in the line (i), then we get- \(\frac{-\lambda \cos ^2 \alpha}{2}-\frac{\lambda \sin ^2 \alpha}{2}=\mathrm{p}\) \(\Rightarrow \frac{-\lambda}{2}=\mathrm{p}\) \(\Rightarrow \lambda=-2 \mathrm{p}\)
TS EAMCET-11.09.2020
Conic Section
119710
If the poles of the line \(x-y=0\) with respect the circles \(x^2+y^2-2 g_i x+c_i^2=0(i=1,2,3)\) are \(\left(\alpha_i\right.\), \(\left.\beta_i\right)\), then \(\sum_{i=1}^3 \frac{\alpha_i+\beta_i}{g_i}=\)
1 3
2 6
3 \(\frac{3}{2}\)
4 \(\frac{3}{4}\)
Explanation:
A Since equation polar of point \(\left(\alpha_i, \beta_i\right)\) with respect to the circle \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{~g}_{\mathrm{i}} \mathrm{x}+\mathrm{c}_{\mathrm{i}}^2=0\) is \(\alpha_i x+\beta_i y-g_i\left(x+\alpha_i\right)+c_i^2=0\) \(\left(\alpha_i-g_i\right) x+\beta_i y+c_i^2-\alpha_i g_i=0\) Now, on comparing with line \(x-y=0\), we get \(\frac{\alpha_{\mathrm{i}}-\mathrm{g}_{\mathrm{i}}}{1}=\frac{\beta_{\mathrm{i}}}{-1}=\frac{\mathrm{c}_{\mathrm{i}}^2-\alpha_{\mathrm{i}} \mathrm{g}_{\mathrm{i}}}{0}\) \(\mathrm{c}_{\mathrm{i}}^2=\alpha_{\mathrm{i}} \mathrm{g}_{\mathrm{i}} \text { and } \alpha_{\mathrm{i}}+\beta_{\mathrm{i}}=\mathrm{g}_{\mathrm{i}}\) \(\therefore \quad \sum_{\mathrm{i}=1}^3 \frac{\alpha_{\mathrm{i}}+\beta_{\mathrm{i}}}{\mathrm{g}_{\mathrm{i}}}=\sum_{\mathrm{i}=1}^3(1)=3\)
TS EAMCET-11.09.2020
Conic Section
119711
If the circle \(x^2+y^2=a^2\) intersects the hyperbola \(x y=b^2\) at four points \(\left(x_1, y_1\right),\left(x_2, y_2\right),\left(x_3, y_3\right)\), \(\left(\mathbf{x}_4, \mathbf{y}_4\right)\), then \(\mathbf{y}_1 \mathbf{y}_2 \mathbf{y}_3 \mathbf{y}_4=\)
1 a
2 0
3 \(b^4\)
4 \(b^2\)
Explanation:
C We have, And \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2\) \(\mathrm{xy}=\mathrm{b}^2\) From Eq. (ii) \(x=\frac{b^2}{y}\) From Eq. (i) \(\left(\frac{\mathrm{b}^2}{\mathrm{y}}\right)^2+\mathrm{y}^2=\mathrm{a}^2\) \(b^4+y^4=a^2 y^2\) \(y^4-a^2 y^2+b^4=0\)This is an equation of 4th degree in \(y\) and its four roots are \(\mathrm{y}_1, \mathrm{y}_2, \mathrm{y}_3, \mathrm{y}_4\) then \(\mathrm{y}_1 \mathrm{y}_2 \mathrm{y}_3 \mathrm{y}_4=\mathrm{b}^4\)
TS EAMCET-10.09.2020
Conic Section
119712
If \(\alpha \neq-4\) and \((2, \alpha)\) is the mid-point of a chord of the circle \(x^2+y^2-4 x+8 y+6=0\), then the values of the \(y\)-intercept of the chord lie in the interval
1 \((-4-\sqrt{14},-4+\sqrt{14})\)
2 \((-4,4)\)
3 \((4, \sqrt{14}, 4+\sqrt{14})\)
4 \((-2,2)\)
Explanation:
A We have, \((2, \alpha)\) is mid-point of chord of circle \(x^2+y^2-4 x+8 y+6=0\) \(\because(2, \alpha)\) inside the circle \(\therefore 4+\alpha^2-8+8 \alpha+6 \leq 0\) \(\alpha^2+8 \alpha+2 \leq 0\) \(\alpha=\frac{-8 \pm \sqrt{64-8}}{2}\) \(\alpha=\frac{-8 \pm 2 \sqrt{14}}{2}=-4 \pm \sqrt{14}\) \(\alpha \in(-4-\sqrt{14},-4+\sqrt{14})\)
119709
If \(x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0\) is the smallest circle through the points of intersection of \(x^2+y^2=a^2\) and \(x \cos \alpha+y \sin \alpha=\) \(\mathbf{p}, \mathbf{0}\lt \mathbf{p}\lt \mathbf{a}\), then \(\lambda=\)
1 1
2 \(-\mathrm{p}\)
3 \(-2 \mathrm{p}\)
4 \(-3 \mathrm{p}\)
Explanation:
C Equation of circle \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{a}^2+\lambda(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha-\mathrm{p})=0\) is the smallest circle, then centre \(\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)\) lies on the line \(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{p}\) \(\therefore\) Put centre in the line (i), then we get- \(\frac{-\lambda \cos ^2 \alpha}{2}-\frac{\lambda \sin ^2 \alpha}{2}=\mathrm{p}\) \(\Rightarrow \frac{-\lambda}{2}=\mathrm{p}\) \(\Rightarrow \lambda=-2 \mathrm{p}\)
TS EAMCET-11.09.2020
Conic Section
119710
If the poles of the line \(x-y=0\) with respect the circles \(x^2+y^2-2 g_i x+c_i^2=0(i=1,2,3)\) are \(\left(\alpha_i\right.\), \(\left.\beta_i\right)\), then \(\sum_{i=1}^3 \frac{\alpha_i+\beta_i}{g_i}=\)
1 3
2 6
3 \(\frac{3}{2}\)
4 \(\frac{3}{4}\)
Explanation:
A Since equation polar of point \(\left(\alpha_i, \beta_i\right)\) with respect to the circle \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{~g}_{\mathrm{i}} \mathrm{x}+\mathrm{c}_{\mathrm{i}}^2=0\) is \(\alpha_i x+\beta_i y-g_i\left(x+\alpha_i\right)+c_i^2=0\) \(\left(\alpha_i-g_i\right) x+\beta_i y+c_i^2-\alpha_i g_i=0\) Now, on comparing with line \(x-y=0\), we get \(\frac{\alpha_{\mathrm{i}}-\mathrm{g}_{\mathrm{i}}}{1}=\frac{\beta_{\mathrm{i}}}{-1}=\frac{\mathrm{c}_{\mathrm{i}}^2-\alpha_{\mathrm{i}} \mathrm{g}_{\mathrm{i}}}{0}\) \(\mathrm{c}_{\mathrm{i}}^2=\alpha_{\mathrm{i}} \mathrm{g}_{\mathrm{i}} \text { and } \alpha_{\mathrm{i}}+\beta_{\mathrm{i}}=\mathrm{g}_{\mathrm{i}}\) \(\therefore \quad \sum_{\mathrm{i}=1}^3 \frac{\alpha_{\mathrm{i}}+\beta_{\mathrm{i}}}{\mathrm{g}_{\mathrm{i}}}=\sum_{\mathrm{i}=1}^3(1)=3\)
TS EAMCET-11.09.2020
Conic Section
119711
If the circle \(x^2+y^2=a^2\) intersects the hyperbola \(x y=b^2\) at four points \(\left(x_1, y_1\right),\left(x_2, y_2\right),\left(x_3, y_3\right)\), \(\left(\mathbf{x}_4, \mathbf{y}_4\right)\), then \(\mathbf{y}_1 \mathbf{y}_2 \mathbf{y}_3 \mathbf{y}_4=\)
1 a
2 0
3 \(b^4\)
4 \(b^2\)
Explanation:
C We have, And \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2\) \(\mathrm{xy}=\mathrm{b}^2\) From Eq. (ii) \(x=\frac{b^2}{y}\) From Eq. (i) \(\left(\frac{\mathrm{b}^2}{\mathrm{y}}\right)^2+\mathrm{y}^2=\mathrm{a}^2\) \(b^4+y^4=a^2 y^2\) \(y^4-a^2 y^2+b^4=0\)This is an equation of 4th degree in \(y\) and its four roots are \(\mathrm{y}_1, \mathrm{y}_2, \mathrm{y}_3, \mathrm{y}_4\) then \(\mathrm{y}_1 \mathrm{y}_2 \mathrm{y}_3 \mathrm{y}_4=\mathrm{b}^4\)
TS EAMCET-10.09.2020
Conic Section
119712
If \(\alpha \neq-4\) and \((2, \alpha)\) is the mid-point of a chord of the circle \(x^2+y^2-4 x+8 y+6=0\), then the values of the \(y\)-intercept of the chord lie in the interval
1 \((-4-\sqrt{14},-4+\sqrt{14})\)
2 \((-4,4)\)
3 \((4, \sqrt{14}, 4+\sqrt{14})\)
4 \((-2,2)\)
Explanation:
A We have, \((2, \alpha)\) is mid-point of chord of circle \(x^2+y^2-4 x+8 y+6=0\) \(\because(2, \alpha)\) inside the circle \(\therefore 4+\alpha^2-8+8 \alpha+6 \leq 0\) \(\alpha^2+8 \alpha+2 \leq 0\) \(\alpha=\frac{-8 \pm \sqrt{64-8}}{2}\) \(\alpha=\frac{-8 \pm 2 \sqrt{14}}{2}=-4 \pm \sqrt{14}\) \(\alpha \in(-4-\sqrt{14},-4+\sqrt{14})\)
119709
If \(x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0\) is the smallest circle through the points of intersection of \(x^2+y^2=a^2\) and \(x \cos \alpha+y \sin \alpha=\) \(\mathbf{p}, \mathbf{0}\lt \mathbf{p}\lt \mathbf{a}\), then \(\lambda=\)
1 1
2 \(-\mathrm{p}\)
3 \(-2 \mathrm{p}\)
4 \(-3 \mathrm{p}\)
Explanation:
C Equation of circle \(\mathrm{x}^2+\mathrm{y}^2-\mathrm{a}^2+\lambda(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha-\mathrm{p})=0\) is the smallest circle, then centre \(\left(\frac{-\lambda \cos \alpha}{2}, \frac{-\lambda \sin \alpha}{2}\right)\) lies on the line \(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{p}\) \(\therefore\) Put centre in the line (i), then we get- \(\frac{-\lambda \cos ^2 \alpha}{2}-\frac{\lambda \sin ^2 \alpha}{2}=\mathrm{p}\) \(\Rightarrow \frac{-\lambda}{2}=\mathrm{p}\) \(\Rightarrow \lambda=-2 \mathrm{p}\)
TS EAMCET-11.09.2020
Conic Section
119710
If the poles of the line \(x-y=0\) with respect the circles \(x^2+y^2-2 g_i x+c_i^2=0(i=1,2,3)\) are \(\left(\alpha_i\right.\), \(\left.\beta_i\right)\), then \(\sum_{i=1}^3 \frac{\alpha_i+\beta_i}{g_i}=\)
1 3
2 6
3 \(\frac{3}{2}\)
4 \(\frac{3}{4}\)
Explanation:
A Since equation polar of point \(\left(\alpha_i, \beta_i\right)\) with respect to the circle \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{~g}_{\mathrm{i}} \mathrm{x}+\mathrm{c}_{\mathrm{i}}^2=0\) is \(\alpha_i x+\beta_i y-g_i\left(x+\alpha_i\right)+c_i^2=0\) \(\left(\alpha_i-g_i\right) x+\beta_i y+c_i^2-\alpha_i g_i=0\) Now, on comparing with line \(x-y=0\), we get \(\frac{\alpha_{\mathrm{i}}-\mathrm{g}_{\mathrm{i}}}{1}=\frac{\beta_{\mathrm{i}}}{-1}=\frac{\mathrm{c}_{\mathrm{i}}^2-\alpha_{\mathrm{i}} \mathrm{g}_{\mathrm{i}}}{0}\) \(\mathrm{c}_{\mathrm{i}}^2=\alpha_{\mathrm{i}} \mathrm{g}_{\mathrm{i}} \text { and } \alpha_{\mathrm{i}}+\beta_{\mathrm{i}}=\mathrm{g}_{\mathrm{i}}\) \(\therefore \quad \sum_{\mathrm{i}=1}^3 \frac{\alpha_{\mathrm{i}}+\beta_{\mathrm{i}}}{\mathrm{g}_{\mathrm{i}}}=\sum_{\mathrm{i}=1}^3(1)=3\)
TS EAMCET-11.09.2020
Conic Section
119711
If the circle \(x^2+y^2=a^2\) intersects the hyperbola \(x y=b^2\) at four points \(\left(x_1, y_1\right),\left(x_2, y_2\right),\left(x_3, y_3\right)\), \(\left(\mathbf{x}_4, \mathbf{y}_4\right)\), then \(\mathbf{y}_1 \mathbf{y}_2 \mathbf{y}_3 \mathbf{y}_4=\)
1 a
2 0
3 \(b^4\)
4 \(b^2\)
Explanation:
C We have, And \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2\) \(\mathrm{xy}=\mathrm{b}^2\) From Eq. (ii) \(x=\frac{b^2}{y}\) From Eq. (i) \(\left(\frac{\mathrm{b}^2}{\mathrm{y}}\right)^2+\mathrm{y}^2=\mathrm{a}^2\) \(b^4+y^4=a^2 y^2\) \(y^4-a^2 y^2+b^4=0\)This is an equation of 4th degree in \(y\) and its four roots are \(\mathrm{y}_1, \mathrm{y}_2, \mathrm{y}_3, \mathrm{y}_4\) then \(\mathrm{y}_1 \mathrm{y}_2 \mathrm{y}_3 \mathrm{y}_4=\mathrm{b}^4\)
TS EAMCET-10.09.2020
Conic Section
119712
If \(\alpha \neq-4\) and \((2, \alpha)\) is the mid-point of a chord of the circle \(x^2+y^2-4 x+8 y+6=0\), then the values of the \(y\)-intercept of the chord lie in the interval
1 \((-4-\sqrt{14},-4+\sqrt{14})\)
2 \((-4,4)\)
3 \((4, \sqrt{14}, 4+\sqrt{14})\)
4 \((-2,2)\)
Explanation:
A We have, \((2, \alpha)\) is mid-point of chord of circle \(x^2+y^2-4 x+8 y+6=0\) \(\because(2, \alpha)\) inside the circle \(\therefore 4+\alpha^2-8+8 \alpha+6 \leq 0\) \(\alpha^2+8 \alpha+2 \leq 0\) \(\alpha=\frac{-8 \pm \sqrt{64-8}}{2}\) \(\alpha=\frac{-8 \pm 2 \sqrt{14}}{2}=-4 \pm \sqrt{14}\) \(\alpha \in(-4-\sqrt{14},-4+\sqrt{14})\)
