If the point (1, 4) lies inside the circle x^ | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

The equation of circle is 

${x^2} + {y^2} - 6x - 10y + P = 0\,\,\,\,\,......\left( i \right)$

${\left( {x - 3} \right)^2} + {\left( {y - 5}

Vedclass · Accepted Answer

$(25, 29)$

Vedclass · Answer

The equation of circle is ${x^2} + {y^2} - 6x - 10y + P = 0\,\,\,\,\,......\left( i \right)$ ${\left( {x - 3} \right)^2} + {\left( {y - 5} \right)^2} = {\left( {\sqrt {34 - P} } \right)^2}$ center $(3,5)$ and radius $'r' = {\left( {\sqrt {34 - P} } \right)^2}$ If cicrle does not touch or intersect the $x$-axis then radium $x < y$ -coordiante of center $C$ or $\sqrt {34 - P} < 5$ $ \Rightarrow 34 - P < 5$ $ \Rightarrow P > 9\,\,\,\,\,\,......\left( {ii} \right)$ Also if the circle does not touch or intersect $X$-axisthe radius $C$ or $\sqrt {34 - P} < 3 \Rightarrow 34 - P < 9 \Rightarrow P > 25\,\,\,\,\,\,......\left( {iii} \right)$ If the point $(1,4)$ is inside the circle, then its distance from center $C < r$ or $\sqrt {\left[ {{{\left( {3 - 1} \right)}^2} + {{\left( {5 - 4} \right)}^2}} \right]} < \sqrt {34 - P} $ $ \Rightarrow 5 < 34 - K$ $ \Rightarrow P < 29\,\,\,\,\,\,\,\,\,\,\,\,\,........\left( {iv} \right)$ Now all the condition $(ii)$, $(ii)$ and $(iv)$ are satisfied if $25 < P < 29$ which is required value of $P$.

If the point $(1, 4)$ lies inside the circle $x^2 + y^2-6x - 10y + p = 0$ and the circle does not touch or intersect the coordinate axes, then the set of all possible values of $p$ is the interval

Similar Questions

Equation of the tangent to the circle, at the point $(1 , -1)$ whose centre is the point of intersection of the straight lines $x - y = 1$ and $2x + y= 3$ is

If $\frac{x}{\alpha } + \frac{y}{\beta } = 1$ touches the circle ${x^2} + {y^2} = {a^2}$, then point $(1/\alpha ,\,1/\beta )$ lies on a/an

The line $(x - a)\cos \alpha + (y - b)$ $\sin \alpha = r$ will be a tangent to the circle ${(x - a)^2} + {(y - b)^2} = {r^2}$

If the tangent to the circle ${x^2} + {y^2} = {r^2}$ at the point $(a, b)$ meets the coordinate axes at the point $A$ and $B$, and $O$ is the origin, then the area of the triangle $OAB$ is

The length of tangent from the point $(5, 1)$ to the circle ${x^2} + {y^2} + 6x - 4y - 3 = 0$, is